The first reaction is a redox reaction, the second and third reactions are none of the above, and the fourth reaction is a precipitation reaction.
Explanation:The reactions asked in the question can be classified as follows:
Ba(ClO₃)₂(s)⟶BaCl₂(s)+3O₂(g): This is a redox reaction. It involves a transfer of electrons which is characterized by changes in oxidation states. Here, chlorine is reduced from +5 in ClO₃⁻ to -1 in Cl⁻, and oxygen is oxidized from -2 in ClO₃⁻ to 0 in O₂.2NaCl(aq)+K₂S(aq)⟶Na₂S(aq)+2KCl(aq): This is a type of double displacement reaction known as 'metathesis', but it can't be classified as acid-base neutralization, redox, or precipitation, so it would fall under none of the above.CaO(s)+CO₂(g)⟶CaCO₃(s): It's a combination reaction resulting in the formation of a single product, calcium carbonate. Given the options, this reaction would also be classified as none of the above.KOH(aq)+AgNO₃(aq)⟶KNO₃(aq)+AgOH(s): This is a precipitation reaction where soluble ions in solution react to form an insoluble product, AgOH(s), which precipitates out of solution.Learn more about Chemical reaction classification here:https://brainly.com/question/8117294
#SPJ3
The first reaction is an oxidation-reduction (combustion) reaction, the second reaction is a precipitation reaction, the third reaction is an acid-base neutralization reaction, and the fourth reaction is also an acid-base neutralization reaction.
Explanation:The first reaction, Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g), is a decomposition reaction also known as oxidation-reduction (combustion). The solid compound breaks down into a solid product and a gas.
The second reaction, 2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq), is a precipitation reaction. The combination of two aqueous solutions forms an insoluble product.
The third reaction, CaO(s)+CO2(g)⟶CaCO3(s), is an acid-base neutralization reaction. The solid oxide reacts with a gas to form a solid carbonate.
The fourth reaction, KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s), is also an acid-base neutralization reaction. The aqueous solutions react to form a solid hydroxide and an aqueous salt.
Learn more about Chemical Reactions here:
https://brainly.com/question/34137415
#SPJ11
Which buffer would be better able to hold a steady pH on the addition of strong acid, buffer 1 or buffer 2? Explain. Buffer 1: a solution containing 0.10 M NH4Cl and 1 M NH3. Buffer 2: a solution containing 1 M NH4Cl and 0.10 M NH3
Answer:
Buffer 1.
Explanation:
Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.
[tex]\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)[/tex].
[tex]\rm NH_3[/tex] gains one hydrogen ion to produce the ammonium ion [tex]\rm {NH_4}^{+}[/tex]. In other words, [tex]\rm {NH_4}^{+}[/tex] is the conjugate acid of the weak base [tex]\rm NH_3[/tex].
Both buffer 1 and 2 include
the weak base ammonia [tex]\rm NH_3[/tex], andthe conjugate acid of the weak base [tex]\rm {NH_4}^{+}[/tex].The ammonia [tex]\rm NH_3[/tex] in the solution will react with hydrogen ions as they are added to the solution:
[tex]\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)[/tex].
There are more [tex]\rm NH_3[/tex] in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of [tex]\rm NH_3[/tex] in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.
A deuteron, 21H, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 3.02×108 K . What is the rms speed of the deuterons? Express your answer using two significant figures.
Answer:
[tex]1.9 \times 10^{6}\text{ m/s}[/tex]
Explanation:
[tex]v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}[/tex]
Data:
T = 3.02 × 10⁸ K
M = 2.013 × 10⁻³ kg/mol
Calculation:
[tex]v_{\text{rms}} = \sqrt{\dfrac{3\times 8.314\text{ J}\cdot\text{K}^{-1} \text{mol}^{-1} \times 3.02 \times 10^{8} \text{ K}}{2.014 \times 10^{-3} \text{ kg}\cdot \text{mol}^{-1}}}\\\\\\=\sqrt{3.740\times 10^{12} \text{ (m/s)}^{2}} = 1.9 \times 10^{6}\text{ m/s}[/tex]
The thermal energy and the conservation of energy allows to find the average speed of the deuteron atoms is:
v = 1.9 10⁶ m / s
The thermal energy of a particle is given by the Boltzmann energy partition relation, which in three dimensions is:
E = [tex]\frac{3}{2}[/tex] kT
Energy kinetic s the energy of movement and its expression is:
K = ½ m v²
They indicate that the temperature of the plasma is T = 3.02 10⁸ K.
If there are no losses, the energy is conserved.
K = E
½ m v² = [tex]\frac{3}{2}[/tex] kT
v = [tex]\sqrt{\frac{3 kT}{m} }[/tex]
[tex]\frac{1.66 \ 10^{-27} kg}{1 u}[/tex]
The mass of deuteron is m = 2.013 u
Let's reduce to kg
m = 2.013 u ( [tex]\frac{1.66 \ 10^{-27} kg}{1 uy}[/tex])
m = 3.5358 10⁻²⁷ kg
We take the mass of a deuterium to 1 mole, multiplying by Avogador's number.
m = 3.5358 10⁻²⁷ 6.022 10²³
m = 2.129 10⁻³ kg / mol
We calculate
v = [tex]\sqrt{ \frac{3 \ 8.314 \ 3.02 \ 10^8 }{2.129 \ 10^{-3} } }[/tex]
v = [tex]\sqrt{3.538 \ 10^{12}}[/tex]
v = 1.88 10⁶ m / s
They ask for the result with two significant figures.
v = 1.9 10⁶ m / s
In conclusion using thermal energy and conservation of energy we can find the average speed of deuteron atoms is:
v = 1.9 10⁶ m / s
Learn more here: brainly.com/question/18989562
A balloon is filled with a gas to a certain volume at a certain pressure at 0.987°C. If the pressure exerted on the balloon is doubled, what must the temperature (in °C) be to keep the balloon inflated at the same volume?
Answer:
275. °CExplanation:
1) Data:
a) Constant volume
b) Initial pressure: P₀
c) Intitial temperature: T₀ = 0.987°C = 0.987 + 273.15 K = 274.137 K
d) Final pressure: P₁ = 2 P₀
e) Final temperature: unknown, T₁
2) Applied principles:
Gay - Lussac's law: at constant volume, the pressure and temperature of the gases are in direct proportion:⇒ P / T = constante ⇒ P₁ / T₁ = P₀ / T₀
3) Solution:
a) Solve for T₁
T₁ = P₁ T₀ / P₀b) Substitute the data:
T₁ = 2P₀ × 274.137 K / P₀ = 548.274 Kc) Converto to °C:
T₁ = 548.274 - 273.15 = 275.124 °C ≈ 275. °C (three significant figures)To keep a balloon inflated at the same volume when the pressure exerted on it is doubled, the temperature must be increased according to Gay-Lussac's Law. After performing the necessary calculations, we found that the new required temperature should be approximately 275.124°C.
Explanation:This question relates to Gas Laws, specifically Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature, assuming the volume stays constant. In this scenario, we need to work out the new temperature when the initial pressure is doubled.
First, let's convert the initial temperature from Celsius to Kelvin. Kelvin = Celsius + 273.15, so, 0.987°C = 274.137K.
According to Gay-Lussac's Law (P₁/T₁ = P₂/T₂ where P is pressure and T is temperature), if P2 = 2*P1 then the new temperature T₂ = 2× T₁, as the volume remains constant.
So, T₂ (in Kelvin) = 2 × T₁ = 2 × 274.137K = 548.274K
To convert T2 back to Celsius, we subtract 273.15 from the Kelvin temperature: T₂ (in Celsius) = 548.274K - 273.15 = 275.124°C.
So, if the pressure exerted on the balloon is doubled, the temperature needs to be approximately 275.124°C to keep the balloon inflated at the same volume.
Learn more about Gas Laws here:https://brainly.com/question/12669509
#SPJ3
Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2(g)+H2O(l)→2HNO3(l)+NO(g) Part A Suppose that 4.2 mol NO2 and 0.50 mol H2O combine and react completely. Which reactant is in excess? Express your answer as a chemical formula. nothing
Final answer:
To determine which reactant is in excess, compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation. In this case, NO2 is in excess.
Explanation:
To determine which reactant is in excess, we need to compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation.
The balanced equation is:
3NO2(g) + H2O(l) -> 2HNO3(l) + NO(g)
According to the equation, the ratio of NO2 to H2O is 3:1. So, for every 3 moles of NO2, we need 1 mole of H2O.
Given that there are 4.2 mol of NO2 and 0.50 mol of H2O, we can calculate the mole ratio of NO2 to H2O:
4.2 mol NO2 / 3 mol NO2 = 1.4 mol NO2
0.50 mol H2O / 1 mol H2O = 0.50 mol H2O
From the calculations, it is clear that the amount of NO2 is more than the required amount based on the stoichiometric ratio. Therefore, NO2 is in excess.
Which statement describes the Arrhenius interpretation of acids and bases?
Answer: the correct answer is option D.<3
Explanation:
e2020
The student is now told that the four solids, in no particular order, are barium chloride (BaCl2), sugar (C6H12O6), butanoic acid (C3H7COOH), and sodium bromide (NaBr). Assuming that conductivity is correlated to the number of ions in solution, rank the four substances based on how well a 0.20 M solution in water will conduct electricity.
Answer:
Explanation:
Out of those choices, the chemical that ionizes the most is NaBr. It is like table salt (NaCl) in its properties. Solutbility 121 grams in 100 mL
You might think "Well if NaBr is such a good conductor, maybe a Metal combined with a non metal is the key and BaCl2 should be next." And in this case you would be right. 36 grams will dissolve in 100 mL
C3H7COOH is an acid and it is soluble in water. It is an acid and the last H splits off. It is not quite as good as the top two, but good enough all the same.
Sugar is probably the least soluble but it does form a suspension. It would be at the bottom of the last.
Answer:
The order of conductivities is: Sugar < Butanoic acid < Sodium bromide < Barium chloride
Explanation:
The conductivity of a solution is related to the concentration of ions. The higher the concentration of ions, the higher the conductivity.
Barium chloride is a strong electrolyte, according to the following equation.
BaCl₂(aq) ⇒ Ba²⁺(aq) + 2 Cl⁻(aq)
Each mole of BaCl₂ produces 3 moles of ions. Thus, if the solution is 0.20 M BaCl₂, it will be 0.60 M in ions.
Sugar (C₆H₁₂O₆) is a non-electrolyte. Thus, the concentration of ions will be zero.
Butanoic acid is a weak electrolyte, according to the following equation.
C₃H₇COOH(aq) ⇄ C₃H₇COO⁻(aq) + H⁺(aq)
Due to its weak nature, the concentration of ions will be lower than 0.20 M.
Sodium bromide is a strong electrolyte, according to the following equation.
NaBr(aq) ⇒ Na⁺(aq) + Br⁻(aq)
Each mole of NaBr produces 2 moles of ions. Thus, if the solution is 0.20 M BaCl₂, it will be 0.40 M in ions.
The order of conductivities is: Sugar < Butanoic acid < Sodium bromide < Barium chloride
For each system listed in first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy For each system listed in first column of the tabl of the system, decrease For each system listed in first column of the tabl , or leave unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column.SystemChange?SA few moles of nitrogen (N2) gas.The nitrogen is compressed from a volume of to a volume of while the temperature is held constant at <?S0=?S0>?S0not enough informationA few grams of liquid acetone ((CH3)2CO).The acetone evaporates at a constant temperature of <?S0=?S0>?S0not enough informationA few moles of nitrogen (N2) gas.The nitrogen is cooled from to and is also expanded from a volume of to a volume of <?S0=?S0>?S0not enough informationFor each system listed in first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy For each system listed in first column of the tabl of the system, decrease For each system listed in first column of the tabl , or leave unchanged. If you don't have enough information to decide, check the "not enough information" button in the last column.SystemChange?SA few moles of nitrogen (N2) gas.The nitrogen is compressed from a volume of to a volume of while the temperature is held constant at <?S0=?S0>?S0not enough informationA few grams of liquid acetone ((CH3)2CO).The acetone evaporates at a constant temperature of <?S0=?S0>?S0not enough informationA few moles of nitrogen (N2) gas.The nitrogen is cooled from to and is also expanded from a volume of to a volume of <?S0=?S0>?S0not enough information
The reason why the entropy of a system increases when it is compressed or when a liquid evaporates is because these processes increase the disorder of the system.
What occurs in the system?When a gas is compressed, the molecules are forced closer together, which means that there are more ways for them to be arranged. Similarly, when a liquid evaporates, the molecules are released from the liquid state and enter the gas state, which means that there are even more ways for them to be arranged.
The reason why the entropy of a system may decrease when it is cooled and expanded is because these processes can decrease the disorder of the system. When a system is cooled, the molecules move more slowly, which means that there are fewer ways for them to be arranged. Similarly, when a system is expanded, the molecules have more space to move around, which also means that there are fewer ways for them to be arranged.
Find out more on entropy here: https://brainly.com/question/30402427
#SPJ2
Both 1,2−dihydronaphthalene and 1,4−dihydronaphthalene may be selectively hydrogenated to 1,2,3,4−tetrahydronaphthalene. One of these isomers has a heat of hydrogenation of 101 kJ/mol (24.1 kcal/mol), and the heat of hydrogenation of the other is 113 kJ/mol (27.1 kcal/mol). Match the heat of hydrogenation with the appropriate dihydronaphthalene.
Answer:
1,4-dihydro = 113 kJ·mol⁻¹
1,2-dihydro = 101 kJ·mol⁻¹
Explanation:
In 1,4-dihydronaphthalene, the 2,3-double bond is isolated from the benzene ring.
In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.
Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).
1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.
The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.
In a constant‑pressure calorimeter, 70.0 mL of 0.350 M Ba(OH)2 was added to 70.0 mL of 0.700 M HCl. The reaction caused the temperature of the solution to rise from 22.32 ∘C to 27.09 ∘C. If the solution has the same density and specific heat as water ( 1.00 g/mL and 4.184J/g⋅K,) respectively), what is Δ???? for this reaction (per mole H2O produced)? Assume that the total volume is the sum of the individual volumes.
Answer:
ΔH = 57 Kj/mole H₂O
Explanation:
Answer:
ΔH/mol H₂O = 55346 J/mol H₂O =55.346 kJ/mol H₂O
Explanation:
The reaction that occurs in this case is:
Ba(OH)₂ + 2 HCl ----> BaCl₂ + 2 H₂O
The measurement and calculation of the amounts of heat exchanged by a system is called calorimetry. The equation that allows calculating these exchanges is:
Q=c*m*ΔT
where Q is the heat exchanged for a body of mass m, constituted by a substance whose specific heat is c, and ΔT is the temperature variation experienced.
In this case:
c=4.184 [tex]\frac{J}{g*C}[/tex] mass solution = mass Ba(OH)₂ + mass HClGiven that the solution has the same density as water (1.00 [tex]\frac{g}{mL}[/tex]) then the mass of Ba (OH)₂ and HCl can be calculated as:
[tex]mass Ba(OH)2=70 mL*\frac{1 g}{1 mL}[/tex]
mass Ba(OH)₂=70 g
[tex]mass HCl=70 mL*\frac{1 g}{1 mL}[/tex]
mass HCl=70 g
mass solution = 70 g of Ba(OH)₂ + 70 g of HCl
mass solution = 140 g
Another way to calculate the mass of the solution is:
The total volume is the sum of the individual volumes:
total volume= volume of Ba(OH)₂ + volume of HCl = 70 mL + 70 mL
total volume= 140 mL
Given that the solution has the same density as water (1.00 [tex]\frac{g}{mL}[/tex]) then
[tex]mass solution=140 mL*\frac{1 g}{1 mL}[/tex]
mass solution = 140 g
ΔT=28.70° C - 24.07 °C= 4.63° CThen
Q = 140 g* 4.184 [tex]\frac{J}{g*C}[/tex] *4.63° C =2712 J
By reaction stochetry (that is, by the relationships between the molecules or elements that make up the reactants and reaction products) 2 moles of HCl produce 2 moles of H2O.
Then
[tex]moles HCl=70 mL*\frac{1 L}{1000 mL} *\frac{0.700 g}{1 L}[/tex]
moles HCl=0.049
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.
If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:
a ⇒ b
c ⇒ x
So
[tex]x=\frac{c*b}{a}[/tex]
In this case: If 2 moles of H2O are formed if 2 moles of HCl react, how many moles of H2O will be formed if 0.049 moles of HCl react?
[tex]moles H2O=\frac{0.049 moles*2moles}{2 moles}[/tex]
moles H2O=0.049 moles
Now
ΔH/mol H₂O = [tex]\frac{2712 J}{0.049 mol H2O}[/tex]
ΔH/mol H₂O = 55346 J/mol H₂O =55.346 kJ/mol H₂O
Classify each of these reactions. 2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq) 2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq) C2H4(g)+3O2(g)⟶2CO2(g)+2H2O(l) C2H4(g)+3O2(g)⟶2CO2(g)+2H2O(l) Cu(s)+FeCl2(aq)⟶Fe(s)+CuCl2(aq) Cu(s)+FeCl2(aq)⟶Fe(s)+CuCl2(aq) Na2S(aq)+2AgNO3(aq)⟶2NaNO3(aq)+Ag2S(s) Na2S(aq)+2AgNO3(aq)⟶2NaNO3(aq)+Ag2S(s)
1. Double replacement but no reaction
2. Double replacement but no reaction
3. Combustion
4. Combustion
5. Single replacement
6. Double replacement
I didn't solve the ones that are repeated! Hope it helped;)
Answer:
1. Double displacement.
2. Combustion.
3. Simple displacement.
4.Double displacement.
Explanation:
Hello,
1. [tex]2HBr(aq)+Ba(OH)_2(aq) -->2H_2O(l)+BaBr_2(aq)[/tex]
In this case, it is about a double displacement reaction since all the cations (H and Ba) and anions (Br and OH) are exchanged.
2. [tex]C_2H_4(g)+3O_2(g)-->2CO_2(g)+2H_2O(l)[/tex]
In this case, it is about the combustion of ethene.
3. [tex]Cu(s)+FeCl_2(aq)-->Fe(s)+CuCl_2(aq)[/tex]
In this case, it is about a simple displacement reaction since the iron (II) cations become solid iron and on the contrary for copper.
4. [tex]Na_2S(aq)+2AgNO_3(aq)-->2NaNO_3(aq)+Ag_2S(s)[/tex]
Finally, it is about a double displacement chemical reaction since the sodium and silver cations are exchanged with the sulfide and nitrate anions.
Best regards.
How many grams of dry NH4Cl need to be added to 1.90 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.72? Kb for ammonia is 1.8×10−5.
Answer : The mass of dry [tex]NH_4Cl[/tex] needed are, 174.196 grams.
Explanation :
First we have to calculate the pOH of the solution.
[tex]pH+pOH=14[/tex]
[tex]pOH=14-pOH\\\\pOH=14-8.72\\\\pOH=5.28[/tex]
Now we have to calculate the [tex]pK_b[/tex].
[tex]pK_b=-\log K_b[/tex]
[tex]pK_b=-\log (1.8\times 10^{-5})[/tex]
[tex]pK_b=4.745[/tex]
Now we have to calculate the concentration of base, [tex]NH_4Cl[/tex].
Using Henderson Hesselbach equation :
[tex]pOH=pKb+\log \frac{[Salt]}{[Base]}[/tex]
[tex]pOH=pKb+\log \frac{[NH_4Cl]}{[NH_3]}[/tex]
Now put all the given values in this equation, we get :
[tex]5.28=4.745+\log \frac{[NH_4Cl]}{0.500}[/tex]
[tex][NH_4Cl]=1.714M[/tex]
Now we have to calculate the mass of [tex]NH_4Cl[/tex].
Formula used :
[tex]Molarity=\frac{\text{Moles of }NH_4Cl}{\text{Volume of solution}}[/tex]
[tex]Molarity=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl\times \text{Volume of solution}}[/tex]
Now put all the given values in this formula, we get:
[tex]1.714M=\frac{\text{Mass of }NH_4Cl}{53.49\times 1.90L}[/tex]
[tex]\text{Mass of }NH_4Cl=174.196g[/tex]
Therefore, the mass of dry [tex]NH_4Cl[/tex] needed are, 174.196 grams.
The pH has been the hydrogen ion concentration and pOH is the hydroxide ion concentration. The mass of ammonium chloride in the solution is 174.196 grams.
What is Kb?The Kb is the base dissociation constant for the compound.
The pH of the solution is 8.72, the pOH of the solution is given as:
[tex]\rm pH=14-pOH\\8.72=14-pOH\\pOH=5.28[/tex]
The pOH of the given solution is 5.28. The concentration of ammonium chloride salt in the solution from pOH can be given as:
[tex]\rm pOH=pKb\;+\;log\dfrac{salt}{acid}[/tex]
The pKb has been given as the logarithmic value of Kb. The concentration of ammonia is 0.5 M. Substituting the values for the concentration of ammonium chloride salt as:
[tex]\rm 5.28=log\;1.8\;\times\;10^{-5}\;+\;log\dfrac{NH_4Cl}{0.5}\\\\ 5.28=4.745\;+\;log\dfrac{NH_4Cl}{0.5}\\\\NH_4Cl=1.714\;M[/tex]
The concentration of ammonium chloride salt is 1.714 M. The volume of the solution is 1.90 L. The molar mass of the compound is 53.49 g/mol.
Substituting the values for the mass of ammonium chloride as:
[tex]\rm Molarity=\dfrac{mass}{molar\;mass\;\times\;volume} \\\\1.714\;M=\dfrac{mass}{53.49\;g/mol\;\times\;1.90\;L}\\\\ Mass=174.196\;g[/tex]
The mass of ammonium chloride added to the solution is 174.196 grams.
Learn more about base dissociation constant, here:
https://brainly.com/question/4256113
Consider the reaction of A(g) + B(g) + C(g) => D(g) for which the following data were obtained:
Experiment Initial [A], mol/L Initial [B], mol/L Initial [C], mol/L Initial rate, mol/L.s
1 0.0500 0.0500 0.0100 6.25 x 10^-3
2 0.100 0.0500 0.0100 2.50 x 10^-2
3 0.100 0.100 0.0100 1.00 x 10^-1
4 0.0500 0.0500 0.0200 6.25 x 10^-3
What is the rate law for the reaction?
Answer:
[tex]r = k [A]^{2}[B]^{2}[/tex]
Explanation:
A + B + C ⟶ D
[tex]\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}[/tex]
Our problem is to determine the values of m, n, and o.
We use the method of initial rates to determine the order of reaction with respect to a component.
(a) Order with respect to A
We must find a pair of experiments in which [A] changes, but [B] and C do not.
They would be Experiments 1 and 2.
[B] and [C] are constant, so only [A] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}[/tex]
(b) Order with respect to B
We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.
They would be Experiments 2 and 3.
[A] and [C] are constant, so only [B] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}[/tex]
(c) Order with respect to C
We must find a pair of experiments in which [C] changes, but [A] and [B] do not.
They would be Experiments 1 and 4.
[A] and [B] are constant, so only [C] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}[/tex]
The rate law for the given reaction is determined to be rate = k[A]^2[B]^2[C]^0, based on careful comparisons of the relationship between initial concentrations of reactants [A], [B], [C] and the observed initial reaction rates in different experiments. The reaction is found to be second order in [A] and [B], and zero order in [C].
Explanation:The rate law for the given chemical reaction can be derived by comparing the changes in the experimentally determined initial rates with changes in the initial concentrations of the reactants [A], [B], and [C]. It's important to note here that the rate law shows how the rate of a reaction depends on the concentration of its reactants.
Looking at Experiments 1 and 2, we observe that the initial concentration of [A] has doubled (from 0.0500 mol/L to 0.100 mol/L), while the initial concentrations of [B] and [C] remain constant. The initial reaction rate quadruples, which suggests that the reaction is second order with respect to [A].
Comparing Experiments 2 and 3, the initial concentration of [B] doubles (from 0.0500 mol/L to 0.100 mol/L), while the concentrations of [A] and [C] remain constant. The rate quadruples, which shows that the reaction order is also second in [B].
Now comparing Experiments 1 and 4, where [C] doubles and [A] and [B] are kept constant, the rate remains unchanged. This suggests the reaction order is zero in [C].
Adding these together, we conclude that the rate law is rate = k[A]^2[B]^2[C]^0, where k is the rate constant.
Learn more about Rate Law here:https://brainly.com/question/35884538
#SPJ6
Show how to convert the temperature 84.7° C to Kelvin. Please include all steps and label the final answer.
Answer:
The answer is 357.85 K.
Explanation:
In order to convert temperatures from degrees Celsius to degrees Kelvin, use the following formula:
temperature in degrees Kelvin = temperature in degrees Celsius + 273.15.
(Oftentimes, 273 is used instead of 273.15; it depends on the discretion of the teacher or student.)
The temperature given in degrees Celsius = 84.7° C
Therefore, the temperature in degrees Kelvin (K) = 84.7 + 273.15 = 357.85 K.
What keeps an inflated balloon from falling down?
Answer:
Helium
Explanation:
If a solution containing 80.701 g80.701 g of mercury(II) perchlorate is allowed to react completely with a solution containing 12.026 g12.026 g of sodium sulfide, how many grams of solid precipitate will form? precipitate: gg How many grams of the reactant in excess will remain after the reaction? excess reactant:
Answer: The excess reagent for the given chemical reaction is mercury (II) perchlorate and the amount left after the completion of reaction is 0.048 moles. The amount of mercury sulfide formed in the reaction is 35.82 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For sodium sulfide:Given mass of sodium sulfide = 12.026 g
Molar mass of sodium sulfide = 78.045 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of sodium sulfide}=\frac{12.026g}{78.045g/mol}=0.154mol[/tex]
For mercury (II) perchlorate:Given mass of mercury (II) perchlorate = 80.701 g
Molar mass of mercury (II) perchlorate = 399.49 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of mercury (II) perchlorate}=\frac{80.701g}{399.49g/mol}=0.202mol[/tex]
For the given chemical equation:
[tex]Hg(ClO_4)_2(aq.)+Na_2S(aq.)\rightarrow HgS(s)+2NaClO_4(aq.)[/tex]
Here, the solid precipitate is mercury sulfide.
By Stoichiometry of the reaction:
1 mole of sodium sulfide reacts with 1 mole of mercury (II) perchlorate.
So, 0.154 moles of sodium sulfide will react with = [tex]\frac{1}{1}\times 0.154=0.154moles[/tex] of mercury (II) perchlorate
As, given amount of mercury (II) perchlorate is more than the required amount. So, it is considered as an excess reagent.
Thus, sodium sulfide is considered as a limiting reagent because it limits the formation of product.
Amount of excess reagent (mercury (II) perchlorate) left = 0.202 - 0.154 = 0.048 molesBy Stoichiometry of the reaction:
1 mole of sodium sulfide reacts with 1 mole of mercury sulfide.
So, 0.154 moles of sodium sulfide will react with = [tex]\frac{1}{1}\times 0.154=0.154moles[/tex] of mercury sulfide.
Now, calculating the mass of mercury sulfide from equation 1, we get:
Molar mass of mercury sulfide = 232.66 g/mol
Moles of mercury sulfide = 0.154 moles
Putting values in equation 1, we get:
[tex]0.154mol=\frac{\text{Mass of mercury sulfide}}{232.66g/mol}\\\\\text{Mass of mercury sulfide}=35.82g[/tex]
Hence, the excess reagent for the given chemical reaction is mercury (II) perchlorate and the amount left after the completion of reaction is 0.048 moles. The amount of mercury sulfide formed in the reaction is 35.82 grams.
We need to find the limiting reactant based on the moles of each reactant in the solution, based on which we can calculate the mass of the solid precipitate that will form and the mass of the reactant in excess that will remain.
Explanation:To solve this problem, we bring in the concept of limiting reactants in a chemical reaction. The limiting reactant is the reactant that is completely consumed in a reaction and determines the maximum amount of product that can be formed. The reaction between mercury(II) perchlorate (Hg(ClO4)2) and sodium sulfide (Na2S) results in the formation of a precipitate. Based on the provided masses, we first need to calculate the moles of each reactant. We'd then compare the stoichiometric ratios to determine the limiting reactant.
After determining the limiting reactant, we can use stoichiometry to calculate how much of the precipitate will form. Furthermore, we can calculate the remaining quantity of the reactant in excess after the reaction.
Please note that the exact calculations require the molecular weights of the reactants and knowledge about the balanced chemical equation of the reaction.
https://brainly.com/question/33417913
#SPJ3
A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 26.6 mL of the base is added. The concentration of acetic acid in the sample was ________ M.
Answer:
.186 M
Explanation:
Remember, the equivalence point is reached when the number of moles of the two reactants is the same.
1. For this problem, obtain the number of moles of NaOH by multiplying the concentration and volume given.
2. Once obtain, the number of moles must be the same for the weak acid.
3. With step two in mind, simply solve for molarity by dividing the value obtained and 25 mL.
Calculation.
1. moles of NaOH = (.175)(.0266) = 4.66e-3
2. moles of Acetic acid = 4.66e-3
3. Concentration of acetic acid: (4.66e-3)/(.025) = .186 M
The concentration of acetic acid in the sample is 0.186 M.
Explanation:The concentration of acetic acid in the sample can be determined using the concept of equivalence point in a titration. From the given information, we know that 26.6 mL of a 0.175 M NaOH solution is needed to reach the equivalence point when titrating a 25.0 mL sample of acetic acid solution.
The equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH → CH3COONa + H2O
Using the balanced equation, we can determine the moles of acetic acid in the sample and then calculate its concentration:
Moles of NaOH = concentration × volume (in liters)
Moles of acetic acid = moles of NaOH
Concentration of acetic acid = moles of acetic acid / volume of acetic acid (in liters)
Substituting the given values, we get:
Concentration of acetic acid = (0.175 M) × (0.0266 L) / (0.025 L) = 0.186 M
If the velocity profile of a fluid emerging from a control surface varies with surface area, the control volume form of the momentum equation can be used. (T/F)
Answer:
False
Explanation:
For different diameter pipes flow rate remains constant but velocity is different. (from continuity A1V1= A2V2). As area changes the velocity of fluid changes
Consider the following reaction at equilibrium: NO2(g) + CO(g) = NO(g) + CO2(g) Suppose the volume of the system is decreased at constant temperature, what change will this cause in the system? A shift to produce more NO A shift to produce more CO A shift to produce more NO2 No shift will occur
Answer: Option (d) is the correct answer.
Explanation:
According to Le Chaltelier's principle, when there occurs any change in an equilibrium reaction then the equilibrium will shift in a direction that will oppose the change.
This means that when pressure is applied on reactant side with more number of moles then the equilibrium will shift on product side that has less number of moles.
For example, [tex]NO_{2}(g) + CO(g) \rightleftharpoons NO(g) + CO_{2}(g)[/tex]
Since here, there are same number of moles on both reactant and product side. So, when volume is decreased at a constant temperature in this system then there will occur no change in the equilibrium state.
Thus, we can conclude that in the given when volume of the system is decreased at constant temperature, then no shift will occur.
Determine the percent sulfuric acid by mass of a 1.61 m aqueous solution of H2SO4. %
Answer:
Wt%H₂SO₄ = 10.2% (w/w)
Explanation:
1.61 molal H₂SO₄ = 1.61 mole H₂SO₄/Kg Solvent = 1.61 mole(98 g/mole)/Kg Solvent = 113.68 g H₂SO₄/Kg Solvent
Solution weight = 1000 g solvent + 113.68 g solute = 1113.68 g Solution
Wt%H₂SO₄ = (113.68 g/1113.68g)100% = 10.2% (w/w)
The percentage of sulfuric acid = 10.2% (w/w)
Given:
Molality of solution = 1.61m
Molality:It is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.
1.61 molal H₂SO₄ = 1.61 mole H₂SO₄/Kg Solvent
= 1.61 mole(98 g/mole)/Kg Solvent = 113.68 g H₂SO₄/Kg Solvent
Solution is made up of solute and solvent. Thus,
Solution weight = 1000 g solvent + 113.68 g solute = 1113.68 g Solution
[tex]Wt\% \text{ of } H_2SO_4 = \frac{113.68 g}{1113.68g}*100\% = 10.2\% (w/w)[/tex]
Find more information about Molality here:
brainly.com/question/9149034
Listed below are the amounts of mercury (in parts per million, or ppm) found in tuna sushi sampled at different stores. The sample mean is 0.836 ppm and the sample standard deviation is 0.253 ppm. Use technology to construct a 90% confidence interval estimate of the mean amount of mercury in the population.
Final answer:
To create a 90% confidence interval for the mean amount of mercury in tuna sushi, use the sample mean (0.836 ppm), sample standard deviation (0.253 ppm), and sample size to calculate the margin of error and apply it to the sample mean.
Explanation:
To construct a 90% confidence interval estimate of the mean amount of mercury in the population, you can use the sample mean (0.836 ppm) and the sample standard deviation (0.253 ppm) along with the sample size to calculate the margin of error. Then add and subtract this margin of error from the sample mean to find the confidence interval.
The formula for the confidence interval is:
Confidence Interval = Sample Mean ± (t * (Sample Standard Deviation / sqrt(Sample Size)))
You will need a t-score for the appropriate degree of freedom and desired confidence level, which you can find using statistical software or a t-distribution table. Once you have the t-score, you can calculate the margin of error and the confidence interval.
This method reflects the uncertainty inherent in estimating a population parameter from a sample statistic and assumes that the sample is representative of the population and that mercury levels are normally distributed.
For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants. 4Al(s)+3O2(g)→2Al2O3(s) A) 1 molAl, 1 mol O2 B) 4 molAl, 2.6 mol O2 C) 16 molAl, 13 mol O2 D) 7.4 molAl, 6.5 mol O2
Answer:
For A: Aluminium is the limiting reagent.
For B: Oxygen gas is the limiting reagent.
For C: Aluminium is the limiting reagent.
For D: Aluminium is the limiting reagent.
Explanation:
Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.
Excess reagent is defined as the reagent which is present in large amount.
For the given chemical reaction:
[tex]4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)[/tex]
For A: 1 mol Al, 1 mol [tex]O_2[/tex]By stoichiometry of the reaction:
4 moles of aluminium reacts with 3 moles of oxygen gas
So, 1 mole of aluminum will react with = [tex]\frac{3}{4}\times 1=0.75moles[/tex] of oxygen gas.
As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.
Hence, aluminium is the limiting reagent.
For B: 4 mol Al, 2.6 mol [tex]O_2[/tex]By stoichiometry of the reaction:
3 moles of oxygen gas reacts with 4 moles of aluminium
So, 2.6 moles of oxygen gas will react with = [tex]\frac{4}{3}\times 2.6=3.458moles[/tex] of aluminium.
As, the given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.
Hence, oxygen gas is the limiting reagent.
For C: 16 mol Al, 13 mol [tex]O_2[/tex]By stoichiometry of the reaction:
4 moles of aluminium reacts with 3 moles of oxygen gas
So, 16 mole of aluminum will react with = [tex]\frac{3}{4}\times 16=12moles[/tex] of oxygen gas.
As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.
Hence, aluminium is the limiting reagent.
For D: 7.4 mol Al, 6.5 mol [tex]O_2[/tex]By stoichiometry of the reaction:
4 moles of aluminium reacts with 3 moles of oxygen gas
So, 7.4 mole of aluminum will react with = [tex]\frac{3}{4}\times 7.4=5.55moles[/tex] of oxygen gas.
As, the given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.
Hence, aluminium is the limiting reagent.
Limiting reactant are defined as the reactant that is involved primarily in the reaction and formation of product depends on this reactant. the limiting reactant for the given equation are as follows:
A: Aluminium is the limiting reagent. B: Oxygen gas is the limiting reagent. C: Aluminium is the limiting reagent. D: Aluminium is the limiting reagent.Limiting reagent are those that limits the formation of products, whereas the reagent present in the excess amount are known as excess reagent.
For A:
For the given chemical reactions, limiting reagents are:
[tex]\text {4 Al}+ \text{3 O}_2\rightarrow\;\text{2Al}_2\text O_3[/tex]By applying the stereochemistry concept:
4 moles of Al reacts with 3 moles of oxygen.
1 mole of Al will react [tex]\dfrac {3}{4}\times 1&= 0.75[/tex] moles of oxygen.Hence, the given amount of oxygen is more than the required amount, it is an excess reagent and aluminum is limiting reagent.
For B:Similarly, in the chemical reaction, where 4 moles of aluminum react with 2.6 moles of oxygen, such that:
3 moles of oxygen reacts with 4 moles of aluminum.
so, 2.6 moles will react with [tex]\dfrac {3}{4}\times 2.6&= 3.458[/tex] moles of aluminum.The given amount of aluminium is more than the required amount, aluminum is considered as an excess reagent and oxygen gas is the limiting reagent.
For C:Also, 16 moles of aluminum reacts with 13 moles of oxygen.
4 moles of aluminum reacts with 3 moles of oxygen, such that:
16 moles of aluminum will react with [tex]\dfrac {3}{4}\times 16&= 12[/tex] moles of oxygen.Thus, oxygen is excess reagent and aluminum is limiting reagent.
For D:7.4 mol Aluminum reacts with 6.5 moles of oxygen, such that:
7.4 moles of aluminum will react with [tex]\dfrac {3}{4}\times 7.4 &= 5.55[/tex] moles of oxygen gas.Therefore, the aluminum is limiting reagent and oxygen is excess reagent.
To know more about limiting reagent, refer to the following link:
https://brainly.com/question/17023950?referrer=searchResults
How is the name of each element in the periodic table represented?
O
A. One or two letters are used to represent the name of each
element.
O
B. Letters are assigned to each element by the scientist who
discovered it.
O
C. Letters are given to represent each element based on when it was
discovered.
O
D. The first two letters of the element's name are used for the table.
Answer:
A. One or two letters are used to represent the name of each element
APEX
One or two letters are used to represent the name of each
element
The name of each element in the periodic table is symbolized by one or two letters.
There are different ways of achieving this. Sometimes;
The first letter of the name of the element is used to represent it, for example, hydrogen = HThe first two letters are sometimes used, for example, Calcium CaThe Latin name is also used for example Gold is Aurum and is represented as Au.These are some of the ways of symbolizing the name of each element.
Learn more about
https://brainly.com/question/20331568
#SPJ2
Given the following equation: 4 NH3 (g)5 O2 (g) >4 NO (g) + 6 H20 (I) How many moles of NH3 is required to react with 25.7 grams of O2?
Answer:
0.6425 moles of [tex]NH_3[/tex] is required to react with 25.7 grams of [tex]O_2[/tex].
Explanation:
mas of oxygen gas = 25.7 g
moles of oxygengas = [tex]\frac{25.7 g}{32 g/mol}=0.8031 mol[/tex]
[tex]4NH_3 (g)+5O_2 (g) \rightarrow 4 NO(g) + 6H_20 (I)[/tex]
According to reaction given above, 5 moles of oxygen gas reacts with 4 moles of ammonia gas.
Then 0.8031 moles of oxygen gas will react with :
[tex]\frac{4}{5}\times 0.8031 mol=0.6425 mol[/tex] of ammonia gas
0.6425 moles of [tex]NH_3[/tex] is required to react with 25.7 grams of [tex]O_2[/tex].
Draw the Lewis structure of PH3. To add lone pairs, click the button before clicking on the molecule. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include lone pairs of electrons and hydrogen atoms. View Available Hint(s)
The Lewis structure of PH₃ is attached to the image below. Phosphine (PH3) consists of a phosphorus atom bonded to three hydrogen atoms.
In the Lewis structure of PH₃, the central phosphorus atom (P) is surrounded by three hydrogen atoms (H). The phosphorus atom has five valence electrons, and each hydrogen atom contributes one valence electron, resulting in a total of eight valence electrons in the structure.
In the lewis structure, each line represents a single bond, and the two dots around the phosphorus atom represent the lone pair of electrons.
To learn more about the Lewis structure, follow the link:
https://brainly.com/question/29603042
#SPJ6
What volume (in mL) of a 0.150 M HNO3 solution will completely react with 35.7 mL of a 0.108 M Na2CO3 solution according to the following balanced chemical equation? Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)
[tex]\frac{number of moles}{Volume}[/tex]Answer:
Explanation:
Given parameters:
Concentration of acid = 0.150M
Volume of base = 35.7mL = 0.0357L
Concentration of base = 0.108M
Unknown:
Volume of acid = ?
The balanced equation of the reactions is given as:
Na₂CO₃ + 2HNO₃ → 2NaNO₃ + CO₂ + H₂O
To find the unkown volume of the acid, we work from the known parameters of the base to the unknown volume of the acid.
Solution
Concentraction is given as:
Molarity = [tex]\frac{Number of moles}{Volume}[/tex]
We first find the number of moles of the base used in the reaction. From the number of moles, we can obtain the volume of the acid used.
Number of moles of base = Molarity x volume of base
= 0.0357 x 0.108 = 0.00386moles
From the balanced reaction equation, we know that:
1 mole of the base reacted with 2 moles of the acid
0.00386 moles of the base would completely react with 0.0077moles
From this, we can now obtain the volume of acid used:
Volume of acid used = [tex]\frac{number of moles of acid}{concentration of acid}[/tex]
Volume of acid = [tex]\frac{0.0077}{0.15}[/tex] = 0.0514L = 51.41mL
The volume in mL of 0.150 M HNO₃ that will completely react with the given Na₂CO₃ is 51.4 mL
To determine the volume of HNO₃ that will completely react with the given Na₂CO₃
From the balanced chemical equation,
Na₂CO₃(aq) + 2HNO₃(aq) → 2NaNO₃(aq) + CO₂(g) + H₂O(l)
This means that 1 mole of Na₂CO₃ reacts with 2 moles of HNO₃ to produce 2 moles of NaNO₃, 1 mole of CO₂ and 1 mole of H₂O
Now, we will determine the number of moles of Na₂CO₃ present in the reaction
From the question,
Volume of Na₂CO₃ = 35.7 mL = 0.0357 L
Concentration of Na₂CO₃ = 0.108 M
From the formula
Number of moles = Concentration × Volume
∴ Number of moles of Na₂CO₃ = 0.108M × 0.0357 L
Number of moles of Na₂CO₃ = 0.0038556 moles
Therefore, 0.0038556 moles of Na₂CO₃ reacted in the reaction
According to the balanced equation,
1 mole of Na₂CO₃ reacts will react completely with 2 moles of HNO₃
∴ 0.0038556 moles of Na₂CO₃ will react with 2 × 0.0038556 moles of HNO₃
Number of moles of HNO₃ = 2 × 0.0038556 moles = 0.0077112 moles
Now, we will determine the volume of 0.150 M HNO₃ that will give this number of moles
From
Number of moles = Concentration × Volume
[tex]Volume = \frac{Number of moles }{Concentration}[/tex]
[tex]Volume = \frac{0.0077112}{0.150}[/tex]
Volume = 0.051408 L = 51.408 mL ≅ 51.4 mL
Hence, the volume in mL of of HNO₃ that will completely react with the given Na₂CO₃ is 51.4 mL
Learn more here: https://brainly.com/question/13394738
Calculate q, w, and ? U for the reversible, isothermal compression of 10 mol of ideal gas from 1 to 10 atm at a constant temperature of 0°C.
Answer : The value of q, w and U for the reversible, isothermal compression are, -52271.69 J, 52271.69 J and 0 J respectively.
Explanation : Given,
Moles of gas = 10 mole
Initial pressure of gas = 1 atm
Final pressure of the gas = 10 atm
Temperature of the gas = [tex]0^oC=273+0=273K[/tex]
According to the question, this is the case of isothermal reversible compression of gas.
As per first law of thermodynamic,
[tex]\Delta U=q+w[/tex]
where,
[tex]\Delta U[/tex] = internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
The expression used for work done will be,
[tex]w=-2.303nRT\log (\frac{P_1}{P_2})[/tex]
where,
w = work done on the gas
n = number of moles of gas
R = gas constant = 8.314 J/mole K
T = temperature of gas
n = moles of the gas
[tex]P_1[/tex] = initial pressure of gas
[tex]P_2[/tex] = final pressure of gas
Now put all the given values in the above formula, we get the work done.
[tex]w=-2.303\times 10mole\times 8.314J/moleK\times 273K\times \log (\frac{1atm}{10atm})[/tex]
[tex]w=52271.69J[/tex]
And we know that, the heat is equal to the work done with opposite sign convention.
So, [tex]q=-52271.69J[/tex]
Therefore, the value of q, w and U for the reversible, isothermal compression are, -52271.69 J, 52271.69 J and 0 J respectively.
A solution is produced in which water is the solvent and there are four solutes. Which of the solutes can dissolve better if the solution is heated?
Answer:
What are the solutes
Explanation:
Which of the following statements about acidic water is true? a. Acid has no effect on the H,O molecules. b. The solution contains a larger number of OH ions than H,O ions. c. The solution contains a larger number of H,O ions than QH ions. d. The solution contains an equal number of H,O ions and OH ions. none of the above e.
Answer:
C
Explanation:
Hydronium (H₃O+) is the same as H+ (aq) because of the same net charge. The acidic property of a solvent or solution is governed the amount of H+ ion in it. The higher the H+ ions the lower the pH. In pure water H₃O+ and OH- are in equal amount. More of OH- ions turn the solution basic.
Write a balanced equation for the combination reaction described, using the smallest possible integer coefficients. When diphosphorus pentoxide combines with water , phosphoric acid is formed.
Answer:
P₂O₅ + 3H₂O => 2H₃PO₄
Explanation:
Nothing to explain.
Write a net ionic equation for the reaction that occurs when excess hydrochloric acid (aq) and potassium sulfite (aq) are combined. Note: Sulfites follow the same solubility trends as sulfates.
Answer: The net ionic equation for the given reaction is [tex]2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow H_2O(l)+SO_2(g)[/tex]
Explanation:
Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:
[tex]2HCl(aq.)+K_2SO_3(aq.)\rightarrow 2KCl(aq.)+SO_2(g)+H_2O(l)[/tex]
Ionic form of the above equation follows:
[tex]2H^+(aq.)+2Cl^-(aq.)+2K^+(aq.)+SO_3^{2-}(aq.)\rightarrow 2K^+(aq.)+2Cl^-(aq.)+SO_2(g)+H_2O(l)[/tex]
As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.
The net ionic equation for the above reaction follows:
[tex]2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow SO_2(g)+H_2O(l)[/tex]
Hence, the net ionic equation for the given reaction is written above.
The net ionic equation for the reaction between hydrochloric acid and potassium sulfite is H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq), following the solubility trends of sulfates and sulfites under standard conditions.
Explanation:The reaction between excess hydrochloric acid (HCl) and potassium sulfite (K2SO3) is a typical acid-base neutralization reaction. In the initial step, potassium sulfite dissociates into its ions in the aqueous solution:
K2SO3 (aq) → 2K+ (aq) + SO3^2- (aq)
Hydrochloric acid, being a strong acid, also dissociates completely:
HCl (aq) → H+ (aq) + Cl- (aq)
The hydrogen ion from the acid then reacts with the sulfite ion to form sulfuric acid and water, creating a net ionic equation :
2H+ (aq) + SO3^2- (aq) → H2SO3 (aq)
Because of the solubility trends of sulfates and sulfites under standard conditions, the sulfuric acid produced also dissociates into ions:
H2SO3 (aq) → 2H+ (aq) + SO3^2- (aq)
Therefore, the overall net ionic equation is:
H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq)
Learn more about Net Ionic Equation here:https://brainly.com/question/35304253
#SPJ3