A. The molarity after a reaction time of 19.00 hours is 0.0988M.
B. 10.12 hours are required for 69% of the [tex][Co(NH_3)_5]Br^{2+}[/tex].
Given data:
Rate constant (k) is [tex]6.3 * 10^{-6} s^{-1}[/tex].
Initial concentration of [tex][Co(NH_3)_5]Br^{2+}[/tex] is 0.100 M.
The rate law for a first-order reaction is R = k * [A]
Where,
R is the rate of the reaction.
k is the rate constant.
[A] is the concentration of the reactant.
A.
To find the molarity after a reaction time of 19.0 hours,
[tex]ln([A]^t / [A]^o) = -kt\\{[A]^t} = [A]^o * e^{-kt}\\{[A]^t} = 0.100 M * e^{-6.3 * 10^{-6} * (19.0 hours * 3600 s/hour)}\\{[A]^t} = 0.0988 M[/tex]
B.
[tex]ln([A]^t / [A]^0) = -kt\\ln(0.69) = -kt[/tex]
Solve for t:
[tex]t = -ln(0.69) / k[/tex]
Substitute the value of k:
[tex]t = -ln(0.69) / (6.3 * 10^{-6} s^{-1})[/tex]
[tex]t = 36441.49 seconds[/tex]
On converting seconds to hours:
t ≈ 10.12 hours
Therefore,
A. The molarity after a reaction time of 19.00 hours is 0.0988M.
B. The time required to react 69% of reactant is 10.12hours.
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The molarity of Co(NH3)5Br2+ after 19 hours is 0.073 M. Approximately 35.75 hours are required for 69% of the Co(NH3)5Br2+ to react.
Explanation:This question pertains to the concept of chemical kinetics, specifically the first order reaction. The molarity after a certain period can be calculated using the formula for first order kinetics: [A] = [A0]e^-kt, where [A0] is the initial concentration, k is the rate constant, t is time, and [A] is the concentration at time t.
A; After 19.0 hours (or 68400 seconds, converted from hours to seconds), the molarity of Co(NH3)5Br2+ can be calculated as follows: [Co(NH3)5Br2+] = (0.100 M)e^-(6.3×10^-6 s^-1 × 68400 s) = 0.073 M.
B; To find the time when 69% of the Co(NH3)5Br2+ has reacted, you need to calculate the time when 31% of the initial molarity remains (as 100% - 69% = 31%). Hence, using the same equation, you can rearrange to find t = -(1/k)ln([A]/[A0]) = -(1/(6.3×10^-6 s^-1)) ln(0.031/0.100) = Approx. 128,680 seconds, or about 35.75 hours.
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The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.42 x 10-4 s-1 at a certain temperature. How long will it take for the concentration of SO2Cl2 to decrease to 25% of its initial concentration
Answer: The time taken is 9764.4 seconds
Explanation:
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]1.42\times 10^{-4}s^{-1}[/tex]
t = time taken for decay process = ? sec
[tex][A_o][/tex] = initial amount of the reactant = 100 grams
[A] = amount left after decay process = 25 grams
Putting values in above equation, we get:
[tex]1.42\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{25}\\\\t=9764.4s[/tex]
Hence, the time taken is 9764.4 seconds
The chemical formula for beryllium sulfide is BeS. A chemist measured the amount of beryllium sulfide produced during an experiment. She finds that 8.31 g of beryllium sulfide is produced. Calculate the number of moles of beryllium sulfide produced. Be sure your answer has the correct number of significant digits
Answer:
The answer is 0.20231 mol BeS
Explanation:
we look for the molecular weight of beryllium and sulfur in the periodic table:
molecular weight of Be = 9.01 g/mol
molecular weight of S = 32.065 g/mol
therefore, the molecular weight of beryllium sulfide is equal to:
molecular weight of BeS = 9.01 + 32.065 = 41.075 g/mol
The number of moles will be equal to:
[tex]n BeS = 8.31 g BeS x\frac{1 mol BeS}{41.075 g BeS}=0.20231 mol BeS[/tex]
Final answer:
To find the number of moles of BeS produced in the experiment, we calculate the molar mass of BeS (41.077 g/mol) and use the formula mass divided by molar mass. The result is 0.202 moles of beryllium sulfide from 8.31 g.
Explanation:
The question asks us to calculate the number of moles of beryllium sulfide (BeS) produced when a chemist measures 8.31 g of it. First, we need to calculate the molar mass of beryllium sulfide. Beryllium (Be) has an atomic mass of 9.012 g/mol, and sulfur (S) has an atomic mass of 32.065 g/mol. Therefore, the molar mass of beryllium sulfide is 9.012 g/mol + 32.065 g/mol = 41.077 g/mol.
To find the number of moles of BeS, we use the formula:
molar mass = mass / number of moles. Rearranging this formula to solve for the number of moles gives number of moles = mass / molar mass. Substituting the given values, we get number of moles = 8.31 g / 41.077 g/mol = 0.202 moles. Therefore, 8.31 g of beryllium sulfide is equivalent to approximately 0.202 moles, with the answer rounded to three significant digits as per the given mass.
The Haber reaction for the manufacture of ammonia is: N2 + 3H2 → 2NH3 Without doing any experiments, which of the following can you say MUST be true? Disappearance rate of H2 = 3 (Disappearance rate of N2). The reaction is first order in N2. Reaction rate = -Δ[N2]/Δt. The reaction is not an elementary reaction. Δ[H2]/Δt will have a positive value. Disappearance rate of N2 = 3 (Disappearance rate of H2). The activation energy is positive.
Final answer:
The Haber process is a reaction between N2 and H2 to form NH3. The disappearance rate of H2 is three times that of N2 (Disappearance rate of H2 = 3 (Disappearance rate of N2)), and the reaction rate can be described as -Δ[N2]/Δt. The activation energy for this reaction must be positive.
Explanation:
The question relates to the Haber process, a reaction between nitrogen (N2) and hydrogen (H2) to form ammonia (NH3). Analyzing the given chemical equation N2 + 3H2 → 2NH3, we can infer several things without any experiments:
For every molecule of nitrogen that reacts, three molecules of hydrogen must also react. Therefore, the disappearance rate of H2 is 3 times the disappearance rate of N2: Disappearance rate of H2 = 3 (Disappearance rate of N2).Reaction rate can be expressed as the negative change in concentration of nitrogen over time: Reaction rate = -Δ[N2]/Δt.Given that the reaction involves a multiple-step process, it is reasonable to say it is not an elementary reaction and thus may not be first order in N2. This cannot be determined without experimental data.The change in concentration of hydrogen over time, Δ[H2]/Δt, will have a negative value because hydrogen is being consumed.Since the reaction requires an input of energy to proceed, the activation energy is positive.Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating the iodination of salicylamide by sodium iodide and sodium hypochlorite via an electrophilic aromatic substitution to form iodo-salicylamide. The density of salicylamide, d = 1.09 g/mL. A reaction was performed in which 3.65 mL of salicylamide was reacted with a mixture of concentrated nitric and sulfuric acids to make 5.33 g of iodosalicylamide. Calculate the theoretical yield and percent yield for this reaction.
Answer: The percent yield of the reaction is 68.68%.
Explanation:
To calculate the mass of salicylamide, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of salicylamide = 1.06 g/mL
Volume of salicylamide = 3.65 mL
Putting values in above equation, we get:
[tex]1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of salicylamide = 3.869 g
Molar mass of salicylamide = 137.14 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol[/tex]
The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:
[tex]\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }[/tex]
By Stoichiometry of the reaction:
1 mole of salicylamide produces 1 mole of iodo-salicylamide
So, 0.0295 moles of salicylamide will produce = [tex]\frac{1}{1}\times 0.0295=0.0295moles[/tex] of iodo-salicylamide
Now, calculating the mass of iodo-salicylamide from equation 1, we get:
Molar mass of iodo-salicylamide = 263 g/mol
Moles of iodo-salicylamide = 0.0295 moles
Putting values in equation 1, we get:
[tex]0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g[/tex]
To calculate the percentage yield of iodo-salicylamide, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of iodo-salicylamide = 5.33 g
Theoretical yield of iodo-salicylamide = 7.76 g
Putting values in above equation, we get:
[tex]\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%[/tex]
Hence, the percent yield of the reaction is 68.68%.
In this case, we can calculate the mass of the salicylamide and use it to calculate the theoretical yield. Then, we can calculate the actual yield and percent yield which is 57.38%
Explanation:To calculate the theoretical yield and percent yield of a reaction, we need to first determine the limiting reagent. From the given information, we know that 3.65 mL of salicylamide was reacted, which has a density of 1.09 g/mL. Therefore, the mass of salicylamide can be calculated as follows:
Mass = Volume x Density = 3.65 mL x 1.09 g/mL = 3.97 g
Next,
Moles of salicylamide used = mass / molar mass = 3.9825 g / 137.14 g/mol = 0.0291 mol
Moles of iodosalicylamide produced = Moles of salicylamide used = 0.0291 mol
Mass of iodosalicylamide produced = moles × molar mass = 0.0291 mol × 319.02 g/mol = 9.303 g
Therefore, the theoretical yield of iodosalicylamide is 9.303 g.
The percentage yield of the reaction can be calculated as follows:
= Actual yield / theoretical yield × 100
= (5.33 g / 9.303 g) × 100 = 57.38%
Therefore, the percentage yield of the reaction is 57.38%.
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Which of the following aqueous solutions are good buffer systems? . 0.29 M perchloric acid + 0.15 M potassium perchlorate 0.16 M potassium acetate + 0.26 M acetic acid 0.18 M hydrofluoric acid + 0.12 M sodium fluoride 0.31 M hypochlorous acid + 0.28 M potassium hypochlorite 0.35 M ammonium bromide + 0.32 M ammonia
Answer:
Good buffer systems are:
Option B) Potassium acetate (KCH3COO) + acetic acid (CH3COOH).
Option C: Hydrofluoric acid (HF) + sodium fluoride (NaF)
Option D: Hypochlorous acid (HClO) + potassium hypochlorite (KClO):
Option E: Ammonium bromide (NH4Br) + ammonia (NH3)
Explanation:
Buffers consist in a mixture of a weak acid with its salt or a weak alkaly with its salt. All buffer systems are conformed by:
1) Weak acid + salt formed with conjugated base of the acid
or
2) Weak alkaly + salt formed with conjugated acid of the alkaly
It is very important these salts come from the weak acid or weak alkaly. It means, the anion of the acid must be the anion in the salt which is going to be part of the buffer system. On the other hand, the cation of the weak alkaly must be the cation of the salt which is going to form the salt in the buffer system.
Then, when we evaluate all options in this exercise, answers are the following:
Option A) Perchloric acid (HClO4) + potassium perchlorate (KClO4).
It is not a buffer system because HClO4 is a strong acid. A buffer requires a weak acid or weak alkaly. KClO4 is a salt formed by the conjugated base of HClO4, but a buffer requires two condition: Weak acid or alkaly + its salt.
Option B) Potassium acetate (KCH3COO) + acetic acid (CH3COOH).
This mixture is a buffer because it is formed by a weak acid (acetic acid) and its salt (KCH3COO is a salt coming from weak acid ---CH3COOH---).
Buffer component reactions:
Reaction weak acid: CH3COOH + H2O <-----> H3O+ + CH3COO-
Reaction salt in water: KCH3COO ---> K+ + CH3COO-
CH3COO- is the anion of the weak acid so it must be part of the salt in the buffer system. Then KCH3COO is a salt from CH3COOH.
Option C: Hydrofluoric acid (HF) + sodium fluoride (NaF)
This mixture is a buffer because it is formed by a weak acid (Hydrofluoric acid) and its salt (NaF is a salt coming from weak acid ---HF---).
Buffer component reactions:
Reaction weak acid: HF + H2O <-----> H3O+ + F-
Reaction salt in water: NaF ---> Na+ + F-
F- is the anion of the weak acid so it must be part of the salt in the buffer system. Then NaF is a salt from HF.
Option D: Hypochlorous acid (HClO) + potassium hypochlorite (KClO):
It is a buffer because it is formed by a weak acid (hypochlorous acid) and its salt (KClO is a salt coming from weak acid ---HClO---).
Buffer component reactions:
Reaction weak acid: HClO + H2O <-----> H3O+ + ClO-
Reaction salt in water: KClO ---> K+ + ClO-
ClO- is the anion of the weak acid so it must be part of the salt in the buffer system. Then KClO is a salt from HClO.
Option E: Ammonium bromide (NH4Br) + ammonia (NH3)
The combination of this weak alkaly (NH3) and the salt (ammonium bromide, NH4Br) is a buffer becuase it is formed by a weak compound and its salt.
Buffer component reactions:
Reaction weak alkaly: NH3 + H2O <-----> NH4+ + OH-
Reaction salt in water: NH4Br ---> NH4+ + OH-
NH4+ is the cation of the weak alkaly so it must be part of the salt in the buffer system. Then NH4Br is a salt from NH3.
Remember a buffer is formed by the combination of two different chemical sustances: Weak acid or Weak alkaly plus its salt.
Final answer:
Good buffer systems consist of a weak acid and its conjugate base or a weak base and its conjugate acid, capable of resisting changes in pH when acids or bases are added. Examples include mixtures of acetic acid with potassium acetate, hydrofluoric acid with sodium fluoride, hypochlorous acid with potassium hypochlorite, and ammonium bromide with ammonia.
Explanation:
Good buffer systems are made of a combination of a weak acid and its conjugate base, or a weak base and its conjugate acid. From the given options, a buffer system can be represented as follows:
0.16 M potassium acetate + 0.26 M acetic acid
0.18 M hydrofluoric acid + 0.12 M sodium fluoride
0.31 M hypochlorous acid + 0.28 M potassium hypochlorite
0.35 M ammonium bromide + 0.32 M ammonia
Each of these systems includes a weak acid or base along with its salt, which is made up of the conjugate base or acid. This combination allows the buffer to resist changes in pH when small amounts of acids or bases are added. For example, when considering acetic acid and sodium acetate as components of the buffer, the addition of a strong acid like HCl will react mainly with the acetate ions rather than causing a large change in pH.
When this system is at equilibrium at a certain temperature PCl5(g) ⇋ PCl3(g) + Cl2(g), the concentrations are found to be [PCl5] = 0.40 M, [PCl3] = [Cl2] = 0.20. If the volume of the container is suddenly halved at the same temperature, what will be the new equilibrium concentration of PCl5?
Answer:
The new concentration of [tex]PCl_5[/tex] will be 0.9 M.
Explanation:
[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]
The equilibrium concentration of all species:
[tex][PCl_5]=0.40 M[/tex]
[tex][PCl_3]=[Cl_2]=0.20 M[/tex]
The equilibrium constant's expression can be written as:
[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
[tex]K_c=\frac{0.20 M\times 0.20 M}{0.40 M}=0.1[/tex]
On halving the volume of the container, the concentration will get doubled;
[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]
0.80 M 0.40M 0.40 M
[tex]Q_c=\frac{0.40 M\times 0.40 M}{0.80 M}=0.2[/tex]
[tex]Q_c>K_c[/tex]
Reaction will go backward.
[tex]PCl_5(g)\rightleftharpoons PCl_3(g) + Cl_2(g)[/tex]
Initially
0.80 M 0.40M 0.40 M
After reestablishment of an equilibrium
(0.80+x) M (0.40-x)M (0.40-x) M
So, the equilibrium expression for above reaction can be written as :
[tex]K_c=\frac{(0.40-x) M\times (0.40-x) M}{(0.80+x) M}[/tex]
[tex]0.1=\frac{(0.40-x) M\times (0.40-x) M}{(0.80+x) M}[/tex]
x = 0.1 M
The new concentration of [tex]PCl_5[/tex] will be:
[tex][PCl_5]=(0.80+x) M=(0.80+0.1) M = 0.90[/tex]
Be sure to answer all parts. In the electrolysis of molten BaI2, (a) What product forms at the negative electrode? (b) What product forms at the positive electrode? You do not need to include the states of matter in your answer.
Answer:
(a) Barium is produced at the negative electrode
(b) Iodine is produced at the positive electrode
Explanation:
When an electric current is passed through a solution containing electrolyte, a non spontaneous reaction is stimulated. This results in the flow of positively charged ions to negatively charged electrodes(cathode) and negatively charged ions to positively charged electrodes(anode)
When an electric current is passed through molten [tex]BaI_{2}[/tex] in the electrolytic cell, the following reactions takes place:
[tex]BaI_{2}[/tex] → [tex]Ba^{2+}[/tex] + 2[tex]I^{-}[/tex]
At the anode;
Iodine ions will lose an electron and will be oxidized to iodine
[tex]2I^{-}[/tex] → [tex]I_{2}[/tex] + [tex]e^{-}[/tex]
At the cathode;
Barium ions gains electrons and its reduced to barium metal
[tex]Ba^{2+}[/tex] + [tex]2e^{-}[/tex] → Ba
In the electrolysis of molten BaI2, Barium forms at the negative electrode, and Iodine forms at the positive electrode.
Explanation:In the electrolysis of molten BaI2, different products form at the negative and positive electrodes. At the negative electrode, also known as the cathode, reduction occurs and Barium (Ba) is formed as a result of Ba2+ ions gaining electrons. On the other hand, at the positive electrode, known as the anode, oxidation happens and Iodine (I2) is formed as a result of I- ions losing electrons.
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A 30.0-mL sample of an unknown strong base is neutralized after the addition of 12.0 mL of a 0.150 M HNO3 solution. If the unknown base concentration is 0.0300 M, give some possible identities for the unknown base.
The question is incomplete, here is the complete question:
A 30.0-mL sample of an unknown strong base is neutralized after the addition of 12.0 mL of a 0.150 M HNO₃ solution. If the unknown base concentration is 0.0300 M, give some possible identities for the unknown base. (Select all that apply)
A.) Ca(OH)₂
B.) LiOH
C.) Sr(OH)₂
D.) Al(OH)₃
E.) NaOH
F.) Ba(OH)₂
Answer: The unknown base could be [tex]Ca(OH)_2,Sr(OH)_2\text{ or }Ba(OH)_2[/tex]
Explanation:
To calculate the number of moles for given molarity of solution, we use the equation:
.......(1)
For nitric acid:Molarity of solution = 0.150 M
Volume of solution = 12.0 mL
Putting values in equation 1, we get:
[tex]0.150M=\frac{\text{Moles of nitric acid}\times 1000}{12.00}\\\\\text{Moles of nitric acid}=\frac{0.150\times 12.00}{1000}=1.8\times 10^{-3}moles[/tex]
For unknown base:Molarity of solution = 0.0300 M
Volume of solution = 30.0 mL
Putting values in equation 1, we get:
[tex]0.0300M=\frac{\text{Moles of unknown base}\times 1000}{30.00}\\\\\text{Moles of unknown base}=\frac{0.0300\times 30.00}{1000}=0.9\times 10^{-3}moles[/tex]
Mole ratio of acid and base is calculated as: [tex]\frac{\text{Moles of unknown base}}{\text{Moles of nitric acid}}=\frac{0.9\times 10^{-3}}{1.8\times 10^{-3}}=\frac{2}{1}[/tex]
Number of [tex]OH^-[/tex] = 2 × number of [tex]H^+[/tex] ions
So, the unknown base is diprotic in nature.
Hence, the unknown base could be [tex]Ca(OH)_2,Sr(OH)_2\text{ or }Ba(OH)_2[/tex]
The unknown base could be a strong base such as sodium hydroxide (NaOH), potassium hydroxide (KOH), barium hydroxide (Ba(OH)2), or calcium hydroxide (Ca(OH)2), as these compounds release two hydroxide ions per molecule.
Explanation:To find the possible identities of the unknown strong base, we first need to understand the neutralization process which involves the reaction of an acid and a base to produce water and salts.
The neutralization reaction is given by: HNO3 + BOH -> H2O + BNO3 where B is the cation from the unknown base. From the stoichiometry of the reaction, we can see that 1 mole of HNO3 reacts with 1 mole of the unknown base.
The number of moles of HNO3 = volume (in L) x molarity = 0.012 L x 0.150 mol/L = 0.0018 mol. So, the number of moles of the unknown base would also be 0.0018 mol.
The molarity of the unknown base = number of moles / volume (in L) = 0.0018 mol / 0.030 L = 0.060 M. But the problem states that the concentration of the unknown base is 0.030 M, which means for every molecule of base, there are two hydroxide ions.
Therefore, the unknown base could be a strong base such as sodium hydroxide (NaOH), potassium hydroxide (KOH), barium hydroxide (Ba(OH)2), or calcium hydroxide (Ca(OH)2), etc., which are all examples of sets of strong bases that release two hydroxide ions per molecule.
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Calculate the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of 78 mL of water from 29 ∘C to 78 ∘C.
To calculate the number of grams of Mg needed for this reaction, use the equation q = mcΔT. By rearranging the equation and substituting the values, you can determine the energy released by the reaction, and then convert it to grams of Mg using the molar mass.
Explanation:To calculate the number of grams of Mg needed for this reaction, we need to use the equation q = mcΔT. We know the initial and final temperatures of the water, as well as the mass and specific heat capacity of water. By rearranging the equation and substituting the values, we can calculate the energy released by the reaction, and then use the molar mass of Mg to determine the number of grams needed.
Let's go through the steps:
Calculate the energy required to increase the temperature of the water using q = mcΔT.Convert the energy to kilojoules by dividing by 1000.Use the molar mass of Mg (24.31 g/mol) to convert the energy to grams of Mg using the equation: grams = energy (kJ) / ΔH (kJ/mol).After performing these calculations, the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of the water from 29 °C to 78 °C is approximately 0.24 g.
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To increase the temperature of 78 mL of water from 29°C to 78°C, 1.36 grams of Magnesium is required given the specific heat capacity of water and the heat of combustion of Magnesium.
Explanation:Firstly, we need to calculate the amount of energy required to heat water from 29°C to 78°C. The specific heat capacity of water is 4.184 J/g°C, and the volume of water is 78 mL, which is equivalent to 78 grams (1 ml of water = 1 gram). To find the energy required, we use the formula Q = m × c × ∆T, where Q is the heat energy required, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature. So,
Q = (78g) × (4.184 J/g°C) × (78°C - 29°C) = 15950.688 Joules
This is the amount of energy we need. Now we need to know how much Mg is needed to produce this energy. Given that Mg reacts to release 11.7 kJ (or 11700 Joules) per gram, we can set up the following equation to solve for mass:
Energy = mass × heat of combustion
Therefore, mass = Energy/Heat of combustion = 15950.688 J / 11700 J/g = 1.36 g
Thus, 1.36 grams of Mg are needed to increase the temperature of 78 mL of water from 29°C to 78°C.
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Iron‑59 is used to study iron metabolism in the spleen. Its half‑life is 44 days. How many days would it take a 28.0 g sample of iron‑59 to decay to 0.875 g?
The radioisotope will take 219 days to decay from 28 g to 0.875 g.
Explanation:
Any radioactive isotope is tend to decay with time. So the rate of decay of the radioactive isotopes is termed as disintegration constant. Since, the initial mass of the radioactive isotope is given along with the reducing mass. In order to determine the time required to reduce the mass of the radioisotope from 28 g to 0.875 g, first the disintegration constant is need to be determined. The disintegration constant can be obtained from half life time of the isotope. As half life time is the measure of time required to reduce half of the concentration of the isotope.
Half life time = 0.6932/disintegration constant
44 = 0.6932/λ
λ = 0.6932/44=0.0158
So, with this values of disintegration constant, initial mass and final mass, the time required to reduce from initial to final mass can be obtained using law of disintegration constant as follows.
N = Noe^(-λt)
[tex]t= -\frac{1}{disintegration constant}* ln(\frac{N}{N_{0} } )\\ \\t = -\frac{1}{0.0158}*ln(\frac{0.875}{28})\\ \\t = -\frac{1}{0.0158}*ln(0.03125)\\\\t = -63.29*(-3.466)\\\\t=219 days.[/tex]
Thus, the radioisotope will take 219 days to decay from 28 g to 0.875 g.
It would take 220 days for 28.0 g sample of iron‑59 to decay to 0.875 g
The half life of a substance is the amount of time it takes to decay to half of the initial value. It is given by:
[tex]N(t)=N_o(\frac{1}{2}) ^\frac{t}{t_\frac{1}{2} } \\\\N(t)=amount\ after\ t\ years, N_o=original\ value,t_\frac{1}{2} =half\ life\\\\Given\ that:\\\\N(t)=0.875g,N_o=28g,t_\frac{1}{2} =44\ days. hence:\\\\\\0.875=28(\frac{1}{2} )^\frac{t}{44} \\\\t=220\ days[/tex]
Hence it would take 220 days for 28.0 g sample of iron‑59 to decay to 0.875 g
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If the final pressure on the gas is 1.20 atm , calculate the entropy change for the process. Check lecture notes or textbook for the formula connecting the change of entropy and gas volume. Assume that neon behaves as an ideal gas in this experiment.
Answer: [tex]-2.34\frac{J}{K}[/tex]
Explanation:
In thermodynamics, entropy (symbolized as S) is a physical quantity for a thermodynamic system in equilibrium. It measures the number of micro states compatible with the macro state of equilibrium, it can also be said that it measures the degree of organization of the system, or that it is the reason for an increase in internal energy versus an increase in the temperature of the system.
If we assume we have 0.6 mol of an ideal gas at 350 K at an initial pressure of 0.75 atm, we calculate the entropy change as:
[tex]S=nRln\frac{P1}{P2}[/tex]
Where S is entropy, n is the number of moles, R is the gas constant R is 8.314 J / mol·K, P1 is the initial pressure and P2 is the final pressure. Then we substitute the values and solve for S.
[tex]S= (0.6 mol).(8.3145\frac{J}{mol.k})(ln\frac{0.75atm}{1.2atm})[/tex]
[tex]S= -2.34\frac{J}{K}[/tex]
The question is not complete and the complete question is ;
the pressure on 0.600 mol of an ideal gas at 350 K is increased isothermally from an initial pressure of 0.750 atm (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process.
Answer:
Entropy change = -2.34 J/k
Explanation:
In thermodynamics, the entropy of a system is expected to increase with increase in temperature, increase in volume, or increase in number of gas particles.
Now, the gas is expanded isothermally and so its temperature is constant. Thus, as pressure of an ideal gas increases, the number of microstates possible for a system decreases and this results in the decrease of reaction entropy and thus is negative.
Thus, ΔS = nRIn(P1/P2)
Now, from the question,
n = 0.06 moles
P1 = 0.75 atm and P1=1.2 atm
R is gas constant and is 8.3145 J/kg.mol
Thus, ΔS = 0.06 x 8.3145 x In(0.75/1.2) = -2.34 J/k
On a balance, you have beakers of AgNO3AgNO3 solution and NaClNaCl solution. When mixed, they will form AgCl(s)AgCl(s). What will happen to the mass of the new mixture?
Explanation:
According to the law of conservation of mass, mass can neither be created nor it can be destroyed. But it can be simply transformed from one form to another.
Therefore, when [tex]AgNO_{3}[/tex] is added to NaCl then the compound formed will have same mass as that of reactants.
[tex]AgNO_{3} + NaCl \rightarrow AgCl + NaNO_{3}[/tex]
Total mass of reactants is (169.87 + 58.44) g/mol = 228.31 g/mol
Total mass of products is (143.32 + 84.99) g/mol = 228.31 g/mol
Thus, we can conclude that mass of the new mixture will stay the same.
At a particular temperature, Kp 0.25 for the reaction a. A ask containing only N2O4 at an initial pressure of 4.5 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A ask containing only NO2 at an initial pressure of 9.0 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and b,does it matter from which direction an equilibrium position is reached?
Answer:
a. pNO₂ = 1 atm pN₂O₄ = 4 atm
b. pNO₂ = 1 atm pN₂O₄ = 4 atm
c. It does not matter.
Explanation:
From the information given in this question we know the equilibrium involved is
N₂O₄ (g) ⇄ 2 NO₂ (g)
with Kp given by
Kp = p NO₂²/ p N₂O₄ = 0.25
We know that if we place 4.5 atm of N₂O₄ is placed in a flask, a quantity x is going to be consumed producing 2x atm of NO₂ and we can setup the following equation:
0.25 = p NO₂²/ p N₂O₄ = (2x)² / (4.5 - x)
0.25 x (4.5 - x) = 4x²
4x² + 0.25 x - 1.125 = 0
after solving this quadratic equation, we get two roots
x₁ = 0.5
x₂ = -0.56
the second root is physically impossible, and the partial pressures for x₁ = 0.5 will be
pNO₂ = 2 x 0.5 atm = 1.0 atm
pN₂O₄ = (4.5 - 0.5) atm = 4.0 atm
Similarly for part b, we get the equilibrium equation
0.25 = (9- 2x)² / x
0.25x = 81 - 36x + 4x²
the roots of this equation are:
x₁ = 5.0625
x₂ = 4
the first root is physically impossible since it will give us a negative partial pressure of N₂O₄ :
p N₂O₄ = 9 - 2(5.0625) = -1.13
the second root give us the following partial pressures:
p N₂O₄ = (9 - 2x4) atm = 1 atm
p NO₂ = 4 atm
The partial pressures are the same, it does not matter from which direction an equilibrium position is reached since what is essential is that the partial pressures of the gasses N₂O₄ and NO₂ obey the equilibrium equation.
Determine if each of the statements is True or False regarding the First Law of Thermodynamics.
1. If the system loses energy to the surroundings, the surroundings could also lose energy to the system.
2. If the system gains thermal energy from the surroundings, the temperature of the surroundings decreases.
3 If the system gains 25 kJ of energy from the surroundings without doing any work on the surroundings, the surroundings could lose 20 kJ of energy.
Answer:
1. False, 2. True, 3. False.
Explanation:
1. False, if the system loses energy to the surroundings, the surroundings receive such energy.
[tex]\Delta U_{sys}<0[/tex]
[tex]\Delta U_{sys} + \Delta U_{surr} = 0[/tex]
[tex]\Delta U_{sys} = -\Delta U_{surr}[/tex]
[tex]\Delta U_{surr}>0[/tex]
2. True, if the system gains more thermal energy, the temperature of the surroundings decreases.
[tex]Q_{sys} = \Delta U_{sys}, Q_{sys}>0\\Q_{surr} =\Delta U_{surr}\\\Delta U_{sys} = -\Delta U_{surr}\\Q_{sys} =-Q_{surr}\\C_{sys} \cdot \Delta T_{sys} = - C_{surr} \cdot \Delta T_{surr}\\\Delta T_{sys}>0,\Delta T_{surr}<0[/tex]
3. False. The surroundings could lose 25 kJ of energy.
1. When the system loses energy, the surrounding should receive the energy. So, it is false.
2. When the system gains thermal energy so the temperature should increase. So, it is true.
3. The surrounding should lose 25 kJ of energy. So, it is false.
First Law of Thermodynamics:a. When the system loses energy, the surrounding should receive the energy. So, it is false.
Like
[tex]\Delta U_{sys} <0\\\\\Delta U_{sys} + \Delta _{surr} = 0\\\\\Delta U_{sys} = -DeltaU_{surr}\\\\DeltaU_{surr} >0[/tex]
b. When the system gains thermal energy so the temperature should increase. So, it is true.
[tex]Q_{sys} = \Delta U_{sys}, Q_{sys}>0\\\\Q_{sys} = \Delta U_{surr}\\\\\Delta U_{sys} = - \Delta U_{surr}\\\\Q_{sys} = -Q{surr}\\\\C_{sys}.\Delta T_{sys} = -C_{surr}.\Delta T_{surr}\\\\\Delta T_{sys}>0,\Delta T_{surr}<0[/tex]
3. The surrounding should lose 25 kJ of energy. So, it is false.
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A buffer solution that is 0.100 M in both HCOOH and HCOOK has a pH = 3.75. A student says that if a very small amount of 0.100 M HCl is added to the buffer, the pH will decrease by a very small amount. Which of the following best supports the student's claim? (A) HCOO will accept a proton from HCl to produce more HCOOH and H2O(B) HCOOH will accept a proton from HCl to produce more HCOO and H2O. (C) HCOO wil donate a proton to HCl to produce more HCOOH and H2O (D) HCOOH will donate a proton to HCl to produce more HCOO and H2O.
Answer: Option (A) is the correct answer.
Explanation:
Chemical equation for the given reaction is as follows.
[tex]HCOO^{-}(aq) + H^{+}(aq) \rightarrow HCOOH(aq)[/tex]
And, the expression to calculate pH of this reaction is as follows.
pH = [tex]pk_{a} + log \frac{[HCOO^{-}]}{[HCOOH]}[/tex]
As the concentration of [tex]HCOO^{-}[/tex] is directly proportional to pH. Hence, when there occurs a decrease in the pH of the solution the [tex][HCOO^{-}][/tex] will also decrease.
Thus, we can conclude that the statement, HCOO will accept a proton from HCl to produce more HCOOH and [tex]H_{2}O[/tex], best supports the student's claim.
The correct option is (A) HCOO will accept a proton from HCl to produce more HCOOH and H.
To understand why option (A) best supports the student's claim, let's consider the buffer system and the effect of adding a small amount of HCl.
A buffer solution consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. In this case, the buffer is composed of formic acid (HCOOH), a weak acid, and its conjugate base, the formate ion (HCOO), along with the potassium salt (HCOOK). The Henderson-Hasselbalch equation relates the pH of the buffer to the concentrations of the weak acid and its conjugate base:
[tex]\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{HCOO}^-]}{[\text{HCOOH}]} \right) \][/tex]
Given that the pH of the buffer is 3.75 and the concentrations of HCOOH and HCOO are both 0.100 M, we can calculate the pKa of formic acid:
[tex]\[ 3.75 = \text{pK}_a + \log \left( \frac{0.100}{0.100} \right) \][/tex]
[tex]\[ 3.75 = \text{pK}_a + \log(1) \][/tex]
[tex]\[ 3.75 = \text{pK}_a \][/tex]
When a small amount of 0.100 M HCl is added to the buffer, the HCl (a strong acid) will dissociate completely into H and Cl ions. The H ions will then react with the formate ions (HCOO) in the buffer, as they are the base component of the buffer system:
[tex]\[ \text{HCOO}^- (aq) + \text{H}^+ (aq) \rightarrow \text{HCOOH} (aq) \[/tex]
This reaction will produce more HCOOH and consume some of the added H ions, thus minimizing the change in pH. The formic acid (HCOOH) will not accept a proton from HCl because it is already protonated; instead, it can only donate a proton, which is not favorable in this case since HCl is a stronger acid.
Option (B) is incorrect because HCOOH cannot accept a proton; it is the acid component of the buffer.
Option (C) is incorrect because HCOO does not donate a proton to HCl; rather, it accepts a proton from HCl.
Option (D) is incorrect because HCOOH does not donate a proton to HCl; HCOOH is a weaker acid than HCl, so it does not donate a proton to HCl's protons.
Therefore, the student's claim that the pH will decrease by a very small amount upon the addition of a small amount of HCl is supported by the fact that the formate ions (HCOO) will accept protons from HCl to produce more formic acid (HCOOH), thus resisting a large change in pH, which is the defining characteristic of a buffer solution.
Problem PageQuestion A chemist fills a reaction vessel with nitrogen monoxide gas, chlorine gas, and nitrosyl chloride gas at a temperature of . Under these conditions, calculate the reaction free energy for the following chemical reaction: Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
The question is incomplete, here is the complete question:
A chemist fills a reaction vessel with 9.47 atm nitrogen monoxide gas, 2.61 atm chlorine gas, and 8.64 atm nitrosyl chloride gas at a temperature of 25°C. Under these conditions, calculate the reaction free energy for the following chemical reaction:
[tex]2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)[/tex]
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
Answer: The Gibbs free energy change of the reaction is -44.0 kJ
Explanation:
The equation used to calculate standard Gibbs free change of a reaction is:
[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}][/tex]
For the given chemical reaction:
[tex]2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)[/tex]
The equation for the standard Gibbs free energy change of the above reaction is:[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NOCl(g))})]-[(2\times \Delta G^o_f_{(NO(g))})+(2\times \Delta G^o_f_{(Cl_2)})][/tex]
We are given:
[tex]\Delta G^o_f_{(NOCl(g))}=66.08kJ/mol\\\Delta G^o_f_{(Cl_2(g))}=0kJ/mol\\\Delta G^o_f_{(NO(g))}=86.55kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=[(2\times (66.08))]-[(2\times (86.55))+(1\times (0))]\\\\\Delta G^o_{rxn}=-40.94kJ[/tex]
The expression of [tex]Q_p[/tex] for above equation follows:[tex]Q_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}[/tex]
We are given:
[tex]p_{NOCl}=8.64atm\\p_{NO}=9.47atm\\p_{Cl_2}=2.61atm[/tex]
Putting values in above expression, we get:
[tex]Q_p=\frac{(8.64)^2}{(9.47)^2\times 2.61}=0.319[/tex]
To calculate the Gibbs free energy change of the reaction, we use the equation:
[tex]\Delta G=\Delta G^o+RT\ln Q_p[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy change = -40.94 kJ
R = Gas constant = 8.314 J/mol.K = 0.008314 kJ/mol.K
T = temperature = [tex]25^oC=[25+273]K=298K[/tex]
[tex]Q_p[/tex] = reaction coefficient = 0.319
Putting values in above equation, we get:
[tex]\Delta G=-40.94+(0.008314J/mol.K\times 298K\times \ln (0.319)\\\\\Delta G=-43.77kJ[/tex]
Hence, the Gibbs free energy change of the reaction is -44.0 kJ
what is the job of a scientist?
A) To ignore facts that do not support his or her theory.
B) To answer ethical questions.
C) To ask and answer scientific questions.
D) To write laws based on his or knowlege.
Answer:
c i believe, not sure
Explanation:
Draw all possible structures for a compound with molecular formula C4H8O that exhibits a broad signal between 3200 and 3600 cm-1 in its IR spectrum and does not contain any signals between 1600 and 1850 cm-1.
The possible structures for C4H8O with an alcohol group and no carbonyl groups, as indicated by the given IR spectrum criteria, include butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol.
Explanation:The IR spectrum criteria given indicate the presence of an alcohol group due to the broad signal between 3200 and 3600 cm⁻¹, which signifies O-H stretching. The absence of signals between 1600 and 1850 cm⁻¹ suggests that there are no carbonyl (C=O) groups in the molecule. Considering these constraints, the molecules of C4H8O that fit these characteristics are butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol. These structures illustrate the versatility of alcohol functionalization within a four-carbon skeleton without the presence of carbonyl groups, thus fitting the IR spectrum restrictions provided.
Which pair of elements do you expect to be most similar in their chemical properties? a. AgAg and KrKr b. OO and SS c. PbPb and OO d. AlAl and CaCa
Answer: Option (b) is the correct answer.
Explanation:
Chemical properties are defined as the properties which tend to chemical composition of a substance. Neutral elements with similar number of valence electrons tend to show similar chemical properties as these have same reactivity which actually affects their chemical properties.
This means that elements of the same group tend to show similar chemical properties.
For example, oxygen and sulfur atom are both group 16 elements and they have 6 valence electrons. Therefore, they show similar chemical properties.
Thus, we can conclude that a pair of O and S elements is expected to be most similar in their chemical properties.
We would like to estimate how quickly the non-uniformities in gas composition in the alveoli are damped out. Consider an alveolus to be spherical with a diameter of 0.1mm. Let the sphere have an initial uniform concentration of oxygen, ci, and at a certain instant, the walls of the alveolus are raised to an oxygen concentration of c¥ and maintained at this value. If the oxygen diffusivity in the alveolus is approximated as that of water, 2.4x10-9 m2 /s, how long does it take for the concentration change (c-ci) at the center to be 90% of the final concentration change?
That duration is 11.775 seconds.
The area of an alveolus, represented as a sphere with a diameter of 0.1 mm, is A = 4.π.r². For an alveolus, r would be: R = 5.10^(-5)m or r = 0.00005m
Locating the region:
4.3.14.(5.10^(-5)) = AA equals 3.14.10^(-8)m³.
Since 90% of the final concentration is to be changed, c = 0.9.3.14.10^(-8)
c = 28.26.10^(-9)
With an oxygen diffusivity of 2.4.10^(-9)m²/s, one alveolus may spread 2.4.10^(-9) oxygen molecules in a second. Thus:
= 2.4.10^(-9)/secondm²t seconds equals (-9)±28.26.10^(28.26.10^(-9) )/(2.4.10^(-9) ) = m² t
t is 11.775 seconds.
It will take 11.775 seconds for the concentration change at the centre to reach 90%.
While mercury is very useful in barometers, mercury vapor is toxic. Given that mercury has a ΔHvap of 59.11 kJ/mol and its normal boiling point is 356.7°C, calculate the vapor pressure in mm Hg at room temperature, 25°C. Group of answer choices 2.99 mm Hg 753 mm Hg 2.68 × 10-3 mm Hg 372 mm Hg
2.38×10^-3
Explanation:
from the question,the we calculate the latent heat of vaporization with the difference in temperature being put into consideration
Four coordinate nickel complexes can have another geometry other than tetrahedral. Name that geometry, and explain why the absorption spectrum you took strongly suggests a tetrahedral geometry, as opposed to the other possible option. Hint: Think about the allowedness of transitions in the two geometries.
Answer:
The other geometry is square planar and this is because the four coordinate complexes forms d⁸ configuration.
Explanation:
(1) The other geometry is square planar
(2) This absorption spectrum (square planar) strongly suggests a tetrahedral geometry because in the molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane.
This square planar is as a results of d⁸ electronic configuration and this is common for transition metal complexes with d⁸ configuration.
Four-coordinate nickel complexes can have either a tetrahedral or square planar geometry. The absorption spectrum suggests a tetrahedral geometry because it allows for more transitions and therefore a more intense color. This is due to the less energy splitting of d orbitals in the tetrahedral geometry.
Explanation:Four-coordinate nickel complexes can adopt either a tetrahedral or a square planar geometry. These different geometries result from the different spatial arrangements of the ligands surrounding the nickel atom.
In a tetrahedral geometry, each of the ligand pairs forms an angle of 109.5°, such as in the complex [Zn(CN)4]²¯. In contrast, square planar complexes, like [Pt(NH3)2Cl2], have the ligands oriented at right angles to each other.
From your absorption spectrum, the presence of a tetrahedral geometry is suggested as opposed to the square planar one due to the allowedness of the transitions. Transition metals with a tetrahedral complex usually allow for more transitions and therefore more bands in the absorption spectrum.
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Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide: PbCO 3 (s) PbO (s) CO 2 (g) ________ grams of lead (II) oxide will be produced by the decomposition of 8.75 g of lead (II) carbonate
Answer:
We will have 7.30 grams lead(II) oxide
Explanation:
Step 1: Data given
Mass of lead (II)carbonate = 8.75 grams
Molar mass PbCO3 = 267.21 g/mol
Step 2: The balanced equation
PbCO3 (s) ⇆ PbO(s) + CO2(g)
Step 3: Calculate moles PbCO3
Moles PbCO3 = mass / molar mass
Moles PbCO3 = 8.75 grams / 267.21 g/mol
Moles PbCO3 = 0.0327 moles
Step 4: Calculate moles PbO
For 1 mol PbCO3 we'll have 1 mol PbO and 1 mol CO2
For 0.0327 moles PbCO3 we'll have 0.0327 moles PbO
Step 5: Calculate mass PbO
Mass PbO = moles PbO * molar mass PbO
Mass PbO = 0.0327 moles * 223.2 g/mol
Mass PbO = 7.30 grams
We will have 7.30 grams lead(II) oxide
Which of the following statements about photosynthesis is correct? (multiple answers may be correct) A: Carbon dioxide serves as final acceptor of high energy electrons provided by the light reactions B: Water serves only as an electron donor in oxygenic photosynthesis C: The oxygen released by oxygenic photosynthesis originates from water D: Hydrogen sulfide can also serve as electron donor in oxygenic photosynthesis E: in anoxygenic photosynthesis carbon dioxide does not serve as final electron acceptor.
Statements B, C, and E are correct about photosynthesis. Water serves as an electron donor in oxygenic photosynthesis and the oxygen released by oxygenic photosynthesis does indeed come from water. In anoxygenic photosynthesis, carbon dioxide is not the end electron acceptor.
Explanation:Among the listed statements about photosynthesis, B: Water serves only as an electron donor in oxygenic photosynthesis, and C: The oxygen released by oxygenic photosynthesis originates from water are correct. During photosynthesis, water is split, releasing electrons and protons, and providing the oxygen gas. In addition, statement E: in anoxygenic photosynthesis carbon dioxide does not serve as final electron acceptor is also correct. In anoxygenic photosynthesis, carried out by certain types of bacteria, something other than water provides the electrons, and carbon dioxide does not serve as the final electron acceptor.
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Statements B, C, and E are correct in regards to photosynthesis. In oxygenic photosynthesis, water acts as an electron donor, and oxygen is released as a byproduct, originating from water. In anoxygenic photosynthesis, carbon dioxide does not serve as the final electron acceptor.
Explanation:Based on the provided information, statements B, C, and E are correct. Let's explain why.
B: In oxygenic photosynthesis, water indeed serves as an electron donor. It helps in replacing the reaction center electron, and as a byproduct, oxygen is formed.
C: The oxygen released by oxygenic photosynthesis indeed originates from water. When water serves as an electron donor and is split, oxygen is released.
E: In anoxygenic photosynthesis, carbon dioxide does not serve as the final electron acceptor. Instead, other reduced molecules like hydrogen sulfide or thiosulfate may be used as the electron donor. Here, oxygen is not formed as a byproduct.
Statements A and D are incorrect. In photosynthesis, Carbon dioxide does not serve as the final acceptor of high energy electrons -- NADP+ does. Besides, Hydrogen sulfide can serve as an electron donor, but not in oxygenic photosynthesis.
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At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M
Answer: Partial pressure of He that would give a solubility of 0.730 M is 15.5 atm
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{He}=K_H\times p_{liquid}[/tex]
where,
[tex]K_H[/tex] = Henry's constant =?
[tex]p_{He}[/tex] = partial pressure = 1.7 atm
Putting values in above equation, we get:
[tex]0.080=K_H\times 1.7atm\\\\K_H=0.047Matm^{-1}[/tex]
To find partial pressure of He would give a solubility of 0.730 M
[tex]0.730=0.047Matm^{-1}\times p_{liquid}[/tex]
[tex]p_{liquid}=15.5atm[/tex]
Thus partial pressure of He that would give a solubility of 0.730 M is 15.5 atm
We have that from the Question, it can be said that The partial pressure of He would give a solubility of 0.730 M is
P_2=4.7atm
From the Question we are told
At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M
Generally the equation for constant temperature is mathematically given as
[tex]\frac{C_2}{C_1}=\frac{P_2}{P_1}\\\\Therefore\\\\P_2=\frac{P_1C_1}{C_1}\\\\P_2=\frac{0.22*1.7}{0.080}\\\\P_2=4.7atm\\\\[/tex]
Therefore
The partial pressure of He would give a solubility of 0.730 M is
P_2=4.7atm
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The compound responsible for the characteristic smell of garlic is allicin, C 6H 10OS 2. The mass of 1.00 mol of allicin, rounded to the nearest integer, is ________ g. 34 162 86 61 19
Answer: The answer is 162g.
Explanation: To calculate the mass of a molecule, you have to find the molar mass of each element of the molecule.
Molar mass of C = 12g/mol
Molar mass of H = 1g/mol
Molar mass of O = 16g/mol
Molar mass of S = 32g/mol
From the allicin's formula, there are:
6 C = 6.12 = 72
10 H = 10.1 = 10
1 O = 1.16 = 16
2 S = 2.32 = 64
Adding them up:
C6H10OS2 = 72+10+16+64 = 162g/mol.
As it is requested 1 mol of allicin, it has 162g.
Answer:
162 g/mol
Explanation:
This problem requires us to calculate the molar mass of allicin, grs/mol. So with the atomic weights of the atoms in this compound, the molar mass will be the sum of the atomic weight multiplied by the number of atoms in the formula.
atomic weights ( g/mol )
C : 12.01 g/mol
H : 1.00 g/mol
O: 16.00 g/mol
S : 32.06 g/mol
Molar mass C₆H₁₀Os₂ (g/mol ) = 6 x 12.01 + 10 x 1.00 + 1 x 16 +2 x 32.06
= 162 g/mol (rounded to the nearest integer)
A 0.10 M imidazole solution has a pH of 6.6. To the nearest hundredth of a unit, what fraction of the molecules are in the neutral (imidazole) form? (The pKa of the imidazolium ion is 6.0.)
Answer : The fraction of the molecules in the neutral (imidazole) form are, 0.799
Explanation : Given,
pH = 6.6
[tex]p_{K_a}=6.0[/tex]
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}[/tex]
Now put all the given values in this expression, we get:
[tex]6.6=6.0+\log \frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}[/tex]
[tex]\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=10^{6.6-6.0}[/tex]
[tex]\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=10^{0.6}[/tex]
[tex]\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=3.98[/tex]
[tex][\text{Imidazole}]=3.98[\text{Imidazolium ion}][/tex] ...........(1)
Now we have to determine the fraction of the molecules are in the neutral (imidazole) form.
Fraction of neutral imidazole = [tex]\frac{[\text{Imidazole}]}{[\text{Imidazole}]+[\text{Imidazolium ion}]}[/tex]
Now put the expression 1 in this expression, we get:
Fraction of neutral imidazole = [tex]\frac{3.98[\text{Imidazolium ion}]}{3.98[\text{Imidazolium ion}]+[\text{Imidazolium ion}]}[/tex]
Fraction of neutral imidazole = [tex]\frac{3.98[\text{Imidazolium ion}]}{4.98[\text{Imidazolium ion}]}[/tex]
Fraction of neutral imidazole = [tex]\frac{3.98}{4.98}[/tex]
Fraction of neutral imidazole = 0.799
Thus, the fraction of the molecules in the neutral (imidazole) form are, 0.799
Final answer:
To find the fraction of imidazole molecules in their neutral form in a 0.10 M solution at pH 6.6, given a pKa of 6.0, we use the Henderson-Hasselbalch equation. The calculation shows that for every 4 molecules of neutral imidazole, there's about 1 deprotonated molecule, demonstrating a significant proportion of neutral molecules in the solution.
Explanation:
The question asks for the fraction of imidazole molecules in the neutral form when a 0.10 M solution of imidazole has a pH of 6.6, given that the pKa of the imidazolium ion is 6.0. This involves understanding the dissociation of acids in solution and applying the Henderson-Hasselbalch equation. The equation is: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the deprotonated form and [HA] is the concentration of the protonated (neutral) form. Solving for the ratio [HA]/[A-] gives us the fraction of molecules in the neutral form.
In this scenario, rearranging the Henderson-Hasselbalch equation and substituting the given values (pH = 6.6 and pKa = 6.0) allows us to solve for the fraction of neutral imidazole molecules: 6.6 = 6.0 + log([HA]/[A-]). This results in a log([HA]/[A-]) of 0.6, which corresponds to a ratio of [HA]/[A-] equal to about 4. Thus, for every 4 molecules of neutral imidazole, there's approximately 1 deprotonated molecule, indicating a large fraction of the molecules remain in the neutral form at pH 6.6.
What is the name of the product of the reaction of 1-pentyne with one equivalent of Br2? (that means that there is one mole of Br2 for each mole of 1-pentyne).
Final answer:
The product of the reaction of 1-pentyne with one equivalent of Br2 is called 1,2-dibromopentene, resulting from a halogen addition reaction.
Explanation:
The reaction of 1-pentyne with one equivalent of Br2 leads to the addition of bromine across the triple bond. Given that there is only one mole of Br2 for each mole of 1-pentyne, the bromine will add to the first and second carbons of the 1-pentyne chain, resulting in a dibromoalkene. Therefore, the name of the product is 1,2-dibromopentene. This product is the result of a typical halogen addition reaction to an alkyne, where the more substituted carbon atom gets the first bromine atom due to the Markovnikov's rule; however, in this case the symmetry of the starting alkyne makes that irrelevant since both terminal carbons are equivalent.
uppose a system consists of 50 mg of organic compound dissolved in 1.00 mL of water (solvent 1). Compare the effectiveness of one 1.5-mL extraction versus three 0.5-mL extractions with ether (solvent 2) where the distribution coefficient, K, is equal to 10. Which is more effective
Answer: the more effective is using three 0.5-mL extractions with ether
Explanation:
Using 1.5 ml of ether
Q = [Vaq/(K×Vorg +Vaq)]^n
Q= remaining fraction
Vaq= volume of aqueous phase
Vorg= volume of organic phase
K= distribution coefficient
Using 1.5 ml ether once
Vaq= 1ml, K= 10, n=1
Q = [Vaq/(K×Vorg +Vaq)]^n
Q= [1/10×1.5+1)]^1 = 0.0625
Using 0.5ml ether 3 times
Q = [1/(10×0.5 +1)]^3= 0.0046
Hence using 0.5ml of ether 3 times is more effective
Consider the following chemical reaction: 2H2O(l)→2H2(g)+O2(g) What mass of H2O is required to form 1.3 L of O2 at a temperature of 295 K and a pressure of 0.926 atm ?
Answer: The mass of water required is 1.79 grams
Explanation:
To calculate the amount of oxygen gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 0.926 atm
V = Volume of the gas = 1.3 L
T = Temperature of the gas = 295 K
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of oxygen gas = ?
Putting values in above equation, we get:
[tex]0.926atm\times 1.3L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 295K\\\\n=\frac{0.926\times 1.3}{0.0821\times 295}=0.0497mol[/tex]
For the given chemical equation:
[tex]2H_2O(l)\rightarrow 2H_2(g)+O_2(g)[/tex]
By Stoichiometry of the reaction:
1 mole of oxygen gas is produced from 2 moles of water
So, 0.0497 moles of oxygen gas will be produced by = [tex]\frac{2}[1}\times 0.0497=0.0994mol[/tex] of water
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of water = 0.0994 moles
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
[tex]0.0994mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.0994mol\times 18g/mol)=1.79g[/tex]
Hence, the mass of water required is 1.79 grams
The mass of H2O required to produce 1.3 L of O2 at 295 K and 0.926 atm can be calculated using the ideal gas law. The volume of O2 is converted to moles using the gas law, which is then doubled to account for the reaction stoichiometry. The result is multiplied by the molar mass of water to find the required mass.
Explanation:To find the mass of H2O required to form 1.3 L of O2 at a temperature of 295 K and a pressure of 0.926 atm, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, calculate the moles of O2 gas using the ideal gas law:
Convert the volume of O2 from liters to cubic meters to match the unit for the gas constant (1 L = 0.001 m³).Use the ideal gas law to calculate the moles of O2.Since the stoichiometry of the reaction 2H2O(l) → 2H2(g) + O2(g) indicates that 2 moles of H2O produce 1 mole of O2, we can double the number of moles of O2 to find the moles of H2O needed.Finally, multiply the moles of H2O by the molar mass of H2O (18.015 g/mol) to find the mass of H2O required.To solve the problem, the calculation steps should be performed with the respective numerical values, which were not provided here because this is only a guideline for how to approach the problem.
Learn more about Ideal Gas Law here:https://brainly.com/question/30458409
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