The balanced equation for the combustion of isooctane (C8H18) is C8H18 + 12.5O2 -> 8CO2 + 9H2O.
Explanation:The balanced equation for the combustion of isooctane (C8H18) to produce carbon dioxide (CO2) and water (H2O) is:
C8H18 + 12.5O2 -> 8CO2 + 9H2O
In this equation, the coefficients have been adjusted to balance the number of atoms on both sides of the equation. There are 8 carbon atoms, 18 hydrogen atoms, and 25 oxygen atoms on both sides of the equation.
Final answer:
The balanced chemical equation for the complete combustion of isooctane (C₈H₁₈) is 2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g), comprising a step-by-step process of first balancing the carbon, then hydrogen, and finally the oxygen atoms.
Explanation:
The balanced chemical equation for the combustion of isooctane (C₈H₁₈) to produce carbon dioxide (CO₂) and water (H₂O) can be represented as follows:
2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g)
Here is a step-by-step explanation:
Start by writing the unbalanced equation: C₈H₁₈ + O₂ → CO₂ + H₂O.Balance the carbon (C) atoms first: As there are 8 C atoms in isooctane, you need to have 8 CO₂ molecules to balance the carbons.Next, balance the hydrogen (H) atoms: With 18 H atoms in isooctane, you will need 9 H₂O molecules.Finally, balance the oxygen (O) atoms: There are 16 O atoms in 8 CO₂ molecules and 18 O atoms in 9 H₂O molecules, giving a total of 25 O₂ molecules needed.After balancing, you end up with the smallest possible integers that balance the equation.
Explain why a fossil record is not 100% accurate.
Snape grows tired of these conceptual questions and thinks it’s time for a problem. What is the retention factor if the distance traveled by the solvent front is 2.00 cm, and the distance traveled by the ion is 0.40 cm?
Answer:
ion travelled : 2.00cm -0.40cm
= 1.6cm
∴ Rf = 1.6/2.0
= 0.80
Explanation:
Retention factor is the ratio of distance travelled by solute divided by distance travelled by solvent.
since you given distance travelled by solvent then the distance required by solute is need in this case the ion.
The Br⊘⊘nsted-Lowry model focuses on the transfer of _______ in an acid-base reaction. The Brnsted-Lowry model focuses on the transfer of _______ in an acid-base reaction. electrons neutrons OH−OH− orbitals H+H+
Answer:
The Brönsted-Lowry model focuses on the transfer of H⁺ in an acid-base reaction.
Explanation:
According to the theory of Brönsted-Lowry , an acid is a chemical substance capable of releasing hydrogen ions, while a base is that chemical substance capable of accepting hydrogen ions. That is, acids are substances capable of yielding protons (hydrogen H + ions) and substance bases capable of accepting them.
On the other hand, the conjugate base of a Brønsted-Lowry acid is the species that forms after an acid donated a proton. The conjugate acid of a Brønsted-Lowry base is the species that forms when a base accepts a proton.
Thus, the acid-base reaction is one in which the acid transfers a proton to a base (proton transfer H⁺).
This theory, unlike another theory like Arrhenius, does not require the presence of water as a means of reaction for the transfer of H⁺.
The Brönsted-Lowry model focuses on the transfer of H⁺ in an acid-base reaction.
Cyclopropane thermally decomposes by a first‑order reaction to form propene. If the rate constant is 9.6 s − 1 , what is the half‑life of the reaction?
Answer:
Half-life of the reaction is 0.072s.
Explanation:
In a first-order reaction, half-life, t1/2, is defined as:
[tex]t_{1/2} = ln2 / k[/tex] (1)
Where k is the rate constant of the reaction.
In the problem, the thermally descomposition of cyclopropane has a rate constant of 9.6s⁻¹. Replacing in (1):
[tex]t_{1/2} = ln2 / 9.6s^{-1}[/tex]
[tex]t_{1/2} = 0.072s[/tex]
I hope it helps!
Your friend, a business major, overhears you talking about chemical energy and food and decides to develop a new super food for busy students that just don’t have time to cook but want to eat a well-balanced diet for a reasonable price. If your friend wants to design a meal bar that contains a high amount of potential energy that the body could use, how would you recommend they rank the possible atom interactions they could incorporate into the chemical formula? Rank from highest chemical energy to lowest.
To design a high energy meal bar, ingredients should be chosen based on their potential chemical energy, with high-energy phosphates, lipids, proteins, and carbohydrates ranked from highest to lowest energy contents.
Ranking Atom Interactions by Chemical Energy
When designing a meal bar with high amounts of potential energy for busy students, your friend should consider the chemical structure of the ingredients. Chemical energy is stored within the bonds of atoms in food molecules. This chemical potential energy is what our bodies metabolize to generate energy, measured in Calories (calorimeter units). Atom interactions in food molecules can be broadly ranked from highest to lowest potential chemical energy as follows:
High-energy phosphates (such as in ATP)
Lipid molecules/fats (like triglycerides)
Protein (amino acid chains)
Carbohydrates (like glucose)
It is important to note that while lipids generally contain more energy per gram than proteins or carbohydrates, the balance and quality of nutrients also play a crucial role in the healthfulness of a meal bar. Therefore, your friend should focus on including a mix of macronutrients that release energy over time and sustain a student's activity levels efficiently.
To create a meal bar with a high amount of potential chemical energy, the macronutrients should be ranked as follows: lipids (fats) with the most chemical energy, followed by proteins, and then carbohydrates.
When designing a superfood meal bar with a high amount of potential chemical energy, we need to consider the atom interactions that store the most energy. The rank of atom interactions from highest to lowest chemical energy typically aligns with the macronutrient composition, as these are the main carriers of chemical energy in food.
Macronutrients Ranked by Chemical Energy:
Lipids (fats) - They have the highest amount of energy per gram due to the presence of long chains of hydrocarbons which, when metabolized, release large amounts of chemical energy.Proteins - Proteins are made up of amino acids linked by peptide bonds, and while they have a considerable energy content, it is lower than that of lipids.Carbohydrates - These include sugars, starches, and fibers. Simple carbohydrates can quickly release energy, while complex carbohydrates provide a more sustained energy release, although both have less energy per gram than lipids and proteins.These rankings are based on the caloric content obtained through oxidation processes like metabolism. In designing a high-energy food bar, your friend should focus on incorporating healthy fats, complete proteins, and complex carbohydrates while considering the overall nutritional balance.
Which of the following solutions would make a good buffer system? (Check all that apply.) A. A solution that is 0.10 M NH3 and 0.10 M NH4Cl B. A solution that is 0.10 M HCN and 0.10 M NaF C. A solution that is 0.10 M HCN and 0.10 M LiCN D. A solution that is 0.10 M HF and 0.10 M NaF
Answer:
A solution that is 0.10 M HCN and 0.10 M LiCN
. A solution that is 0.10 M NH3 and 0.10 M NH4Cl
Explanation:
A buffer consists of a weak acid and its conjugate base counterpart. HCN is a weak acid and the salt LiCN contains its counterpart conjugate base which is the cyanide ion. A buffer maintains the pH by guarding against changes in acidity or alkalinity of the solution.
A solution of ammonium chloride and ammonia will also act as a basic buffer. A buffer may also contain a weak base and its conjugate acid.
Answer:
Good buffer systems are:
A) NH3 + NH4Cl
C) HCN + LiCN
D) HF + NaF
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
or
2) Weak alkaly + salt
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:
A) 0.10 M NH3 and 0.10 M NH4Cl. It is a buffer because NH3 (ammonia) is a weak alkaly and NH4Cl is a salt coming from NH3.
Buffer component reactions:
Reaction weak alkaly: NH3 + H2O <-----> NH4+ + OH-
Reaction salt in water: NH4Cl ---> NH4+ + Cl-
NH4+ is the cation of the weak alkaly so it must be part of the salt in the buffer system. Then NH4Cl is a salt from NH3.
C) 0.10 M HCN and 0.10 M LiCN. It is a buffer because HCN is a weak acid and LiCN is a salt which is coming from HCN.
Buffer component reactions:
Reaction weak acid: HCN + H2O <-----> H3O+ + CN-
Reaction salt in water: LiCN --> Li+ + CN-
CN- is the anion of the acid, so it must be part of the salt in the buffer system. Then LiCN is a salt from HCN.
D) 0.10 M HF and 0.10 M NaF. It is a buffer because HF is a weak acid and NaF is a salt which is coming from 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 (HF), so it must be part of the salt in th buffer systema. Then NaF is a salt coming from HF.
However option B, it is not a buffer, because it is a mixture of 0.10 M HCN and 0.10 M NaF. Salt is not coming from the weak acid.
Reaction weak acid: HCN + H2O <-----> H3O+ + CN- (anion of the acid is CN-)
Rection salt in water: NaF --> Na+ + F- (anion in the salt is F-, not CN-)
Anion of the acid is CN- and the anion in the salt is F- so it is not a salt coming from the weak acid. Then option B it is not a buffer system.
Use the rules (in order) to assign oxidation numbers to each of the elements in the compounds below.
1. hydrogen phosphate ion HPO32-
H P O
2. aluminum oxide Al2O3
Al O
3. periodic acid HIO4
H I O
Answer :
(1) The oxidation number of P is, (+3)
The oxidation number of H is, (+1)
The oxidation number of O is, (-2)
(2) The oxidation number of Al is, (+3)
The oxidation number of O is, (-2)
(3) The oxidation number of I is, (+7)
The oxidation number of H is, (+1)
The oxidation number of O is, (-2)
Explanation :
Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.
When the atoms are present in their elemental state then the oxidation number will be zero.
Rules for Oxidation Numbers :
The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.
The oxidation number of a Group 1 element in a compound is +1.
The oxidation number of a Group 2 element in a compound is +2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
(1) The given compound is, [tex]HPO_3^{2-}[/tex]
Let the oxidation state of 'P' be, 'x'
[tex](+1)+x+3(-2)=-2\\\\x=+3[/tex]
The oxidation number of P is, (+3)
The oxidation number of H is, (+1)
The oxidation number of O is, (-2)
(2) The given compound is, [tex]Al_2O_3[/tex]
Let the oxidation state of 'Al' be, 'x'
[tex]2x+3(-2)=0\\\\x=+3[/tex]
The oxidation number of Al is, (+3)
The oxidation number of O is, (-2)
(3) The given compound is, [tex]HIO_4[/tex]
Let the oxidation state of 'I' be, 'x'
[tex](+1)+x+4(-2)=0\\\\x=+7[/tex]
The oxidation number of I is, (+7)
The oxidation number of H is, (+1)
The oxidation number of O is, (-2)
The rules of determining oxidation numbers have been applied to three compounds: hydrogen phosphate (with oxidation numbers of +1 for H, +5 for P and -2 for O), aluminum oxide (+3 for Al and -2 for O), and periodic acid (+1 for H,+7 for I, and -2 for O).
Explanation:The rules to assign oxidation numbers to the elements in the given compounds may be understood as follows:
For the hydrogen phosphate ion HPO32-, the oxidation number of hydrogen, H, is usually +1. For oxygen, O, it is usually -2. The overall charge of the ion is -2. Let P's oxidation number be x. So, using the rule, x + 1*(1) + 3*(-2) = -2. Solving for x gives x = +5, so phosphorus, P, has an oxidation number of +5.For aluminium oxide, Al2O3, Aluminium Al always has an oxidation state of +3 in its compounds. Oxygen, has an oxidation state of -2. Therefore, for Al2O3, the oxidation states of Al and O are +3 and -2 respectively.Lastly, for periodic acid, HIO4, the oxidation number of H is +1, and for O is -2. Let's let the oxidation number of I (iodine) be represented by x. So, based on the equation x + 1 + 4*(-2) =0, solving for x gives x = +7. Hence, the oxidation state of I is +7.Learn more about oxidation numbers here:https://brainly.com/question/29288807
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The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.
NO2(g) + O3(g) --> NO3 (g) + O2 (g)
If the rate constant for the reaction is 1.69 x 10^-4 M-1s-1 at 298 K. What is the rate of the reaction when [NO2] = 1.77 x 10^-8 M and [O3] = 1.59 x 10-7 M?
________ M/s
What is the rate of the appearance of NO3 under these conditions?
________M/s
Answer : The rate of reaction is,
[tex]Rate=4.77\times 10^{-19}M/s[/tex]
The appearance of [tex]NO_3[/tex] is, [tex]4.77\times 10^{-19}M/s[/tex]
Explanation :
The general rate of reaction is,
[tex]aA+bB\rightarrow cC+dD[/tex]
Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The expression for rate of reaction will be :
[tex]\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}[/tex]
[tex]\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}[/tex]
[tex]\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}[/tex]
[tex]\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]
[tex]Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]
From this we conclude that,
In the rate of reaction, A and B are the reactants and C and D are the products.
a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.
The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.
The given rate of reaction is,
[tex]NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)[/tex]
The rate law expression will be:
[tex]Rate=k[NO_2][O_3][/tex]
Given:
Rate constant = [tex]k=1.69\times 10^{-4}M^{-1}s^{-1}[/tex]
[tex][NO_2][/tex] = [tex]1.77\times 10^{-8}M[/tex]
[tex][O_3][/tex] = [tex]1.59\times 10^{-7}M[/tex]
[tex]Rate=k[NO_2][O_3][/tex]
[tex]Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})[/tex]
[tex]Rate=4.77\times 10^{-19}M/s[/tex]
The expression for rate of appearance of [tex]NO_3[/tex] :
[tex]\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}[/tex]
As, [tex]\text{Rate of reaction}=4.77\times 10^{-19}M/s[/tex]
So, [tex]\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s[/tex]
Thus, the appearance of [tex]NO_3[/tex] is, [tex]4.77\times 10^{-19}M/s[/tex]
The rate of reaction is [tex]4.77*10^{-19} M/s[/tex]
The rate of the appearance of [tex]NO_{3}[/tex] is, [tex]4.77*10^{-19} M/s[/tex]
The second order reaction is given as,
[tex]NO_{2}(g) + O_{3}(g) \Rightarrow NO_{3} (g) + O_{2} (g)[/tex]
Given that,
Rate constant [tex]k=1.69*10^{-4}[/tex]
[tex][NO_{2}] = 1.77 *10^{-8} M\\\\O_{3}= 1.59 * 10^{-7} M[/tex]
Rate of reaction is given as,
[tex]Rate=k[NO_{2}][O_{3}]\\\\Rate=1.69*10^{-4}* 1.77*10^{-8}*1.59*10^{-7}\\\\Rate=4.77*10^{-19} M/s[/tex]
the rate of the appearance of [tex]NO_{3}[/tex] is,
[tex]\frac{d[NO_{3}]}{dt}=4.77*10^{-19} M/s[/tex]
Learn more about the Rate of reaction here:
https://brainly.com/question/23637409
Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27% b. caffeine (found in coffee beans): C 49.48%, H 5.19
This an incomplete question, here is a complete question.
Calculate the empirical formula for each stimulant based on its elemental mass percent composition.
a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%
b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%
Answer:
(a) The empirical formula for the given compound is [tex]C_5H_7N[/tex]
(b) The empirical formula for the given compound is [tex]C_4H_5N_2O[/tex]
Explanation:
Part A: nicotine
We are given:
Percentage of C = 74.03 %
Percentage of H = 8.70 %
Percentage of N = 17.27 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 74.03 g
Mass of H = 8.70 g
Mass of N = 17.27 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles[/tex]
Moles of Nitrogen = [tex]\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.
For Carbon = [tex]\frac{6.17}{1.23}=5.01\approx 5[/tex]
For Hydrogen = [tex]\frac{8.70}{1.23}=7.07\approx 7[/tex]
For Nitrogen = [tex]\frac{1.23}{1.23}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 5 : 7 : 1
The empirical formula for the given compound is [tex]C_5H_7N_1=C_5H_7N[/tex]
Part B: caffeine
We are given:
Percentage of C = 49.48 %
Percentage of H = 5.19 %
Percentage of N = 28.85 %
Percentage of O = 16.48 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 49.48 g
Mass of H = 5.19 g
Mass of N = 28.85 g
Mass of O = 16.48 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles[/tex]
Moles of Nitrogen = [tex]\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.
For Carbon = [tex]\frac{4.12}{1.03}=4[/tex]
For Hydrogen = [tex]\frac{5.19}{1.03}=5.03\approx 5[/tex]
For Nitrogen = [tex]\frac{2.06}{1.03}=2[/tex]
For Nitrogen = [tex]\frac{1.03}{1.03}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N : O = 4 : 5 : 2 : 1
The empirical formula for the given compound is [tex]C_4H_5N_2O_1=C_4H_5N_2O[/tex]
Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g of NaCl, and you isolate 52.1 g of PbCl2, what is your percent yield?
Answer:
[tex]\large \boxed{84.7 \, \%}[/tex]
Explanation:
Mᵣ: 58.44 278.11
Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃
m/g: 26.3
1. Moles of NaCl
[tex]\text{Moles of NaCl} = \text{26.3 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}[/tex]
(b) Moles of PbCl₂
[tex]\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} \times \dfrac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}[/tex]
(c) Theoretical yield of PbCl₂
[tex]\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} \times \dfrac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}[/tex]
(d) Percent yield
[tex]\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \,\% = \dfrac{\text{52.1 g}}{\text{61.52 g}} \times 100 \, \% = \mathbf{84.7 \,\%}\\\\\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}[/tex]
Final answer:
In a reaction where lead (II) nitrate reacts with sodium chloride to form lead (II) chloride, given 26.3 g of NaCl and 52.1 g of PbCl2 produced, the percent yield is calculated to be 83.27%.
Explanation:
The student is performing a reaction where lead (II) nitrate reacts with sodium chloride to produce lead (II) chloride and sodium nitrate. To calculate the percent yield, we use the actual mass of PbCl2 obtained from the experiment (52.1 g), and compare it with the theoretical mass that should have been produced if the reaction were 100% efficient.
First, you need to calculate the moles of NaCl:
Molecular weight of NaCl = 22.99 (Na) + 35.45 (Cl) = 58.44 g/molMoles of NaCl = mass (g) / molar mass (g/mol) = 26.3 g / 58.44 g/mol = 0.45 molesBased on the stoichiometry of the balanced equation, 2 moles of NaCl will produce 1 mole of PbCl2 (2:1 ratio). Therefore, 0.45 moles of NaCl should theoretically produce 0.225 moles of PbCl2.
Now, calculate the theoretical yield:
Molecular weight of PbCl2 = 207.2 (Pb) + 2*35.45 (Cl) = 278.1 g/molTheoretical yield (g) = moles * molar mass = 0.225 moles * 278.1 g/mol = 62.57 g of PbCl2To find the percent yield:
Percent yield = (actual yield / theoretical yield) * 100% = (52.1 g / 62.57 g) * 100% ≈ 83.27%The percent yield of the reaction is therefore 83.27%.
Hydrogen peroxide decomposes to water and oxygen according to the following reaction H 2 O 2(aq) → H 2 O + ½ O 2 (g) It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate (KMnO4) at certain intervals
This is an incomplete question, here is a complete question.
Hydrogen peroxide decomposes to water and oxygen according to the following reaction:
[tex]H_2O_2(aq)\rightarrow H_2O+\frac{1}{2}O_2(g)[/tex]
It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate at certain intervals.
Initial rate determinations at 40°C for the decomposition give the following data:
[H₂O₂] (M) Initial Rate (mol/L min)
0.10 1.93 × 10⁻⁴
0.20 3.86 × 10⁻⁴
0.30 5.79 × 10⁻⁴
Calculate the half-life for the reaction at 40°C?
Answer : The half life for the reaction is, 3590.7 minutes
Explanation :
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
[tex]H_2O_2(aq)\rightarrow H_2O+\frac{1}{2}O_2(g)[/tex]
Rate law expression for the reaction:
[tex]\text{Rate}=k[H_2O_2]^a[/tex]
where,
a = order with respect to [tex]H_2O_2[/tex]
Expression for rate law for first observation:
[tex]1.93\times 10^{-4}=k(0.10)^a[/tex] ....(1)
Expression for rate law for second observation:
[tex]3.86\times 10^{-4}=k(0.20)^a[/tex] ....(2)
Expression for rate law for third observation:
[tex]5.79\times 10^{-4}=k(0.30)^a[/tex] ....(3)
Dividing 2 by 1, we get:
[tex]\frac{3.86\times 10^{-4}}{1.93\times 10^{-4}}=\frac{k(0.20)^a}{k(0.10)^a}\\\\2=2^a\\a=1[/tex]
Thus, the rate law becomes:
[tex]\text{Rate}=k[H_2O_2]^1[/tex]
[tex]\text{Rate}=k[H_2O_2][/tex]
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
[tex]1.93\times 10^{-4}=k(0.10)^1[/tex]
[tex]k=1.93\times 10^{-3}min^{-1}[/tex]
Now we have to calculate the half-life for the reaction.
The expression used is:
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]1.93\times 10^{-3}min^{-1}=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=3590.7min[/tex]
Thus, the half life for the reaction is, 3590.7 minutes
Estimate the following: a) The volume occupied by 18 kg of ethylene at 55°C and 35 bar. b) The mass of ethylene contained in a 0.25-m3 cylinder at 50°C and 115 bar.
The estimated volume occupied by 18 kg of ethylene at 55℃ and 35 bar is approximately 476.6 liters. The mass of ethylene contained in a [tex]0.25m^3[/tex] cylinder at 50℃ and 115 bar is approximately 3.03 kg.
To estimate the volume occupied by 18 kg of ethylene at 55℃ and 35 bar, we can use the ideal gas law expressed as PV = nRT. First, we need to calculate the number of moles (n) using the molar mass of ethylene ([tex]C_2H_4[/tex]) which is approximately 28 g/mol. Therefore, n = 18000 g / 28 g/mol = 642.86 moles. The gas constant (R) is 0.08314 L*bar/mol*K. Converting the temperature to Kelvin, T = 55 + 273 = 328 K. Now, we can solve for V:
V = (nRT) / P = (642.86 mol * 0.08314 L*bar/mol*K * 328 K) / 35 bar = 476.6 L
To determine the mass of ethylene in a 0.25-[tex]m^3[/tex] cylinder at 50℃ and 115 bar, we apply the ideal gas law again. We first convert the volume to liters (V = 0.25 [tex]m^3[/tex] * 1000 L/[tex]m^3[/tex] = 250 L), and temperature to Kelvin (T = 50 + 273 = 323 K). The pressure is already in bar. Now, we calculate the number of moles, solving for n:
n = PV / RT = (115 bar * 250 L) / (0.08314 L*bar/mol*K * 323 K) = 108.14 moles
The mass of ethylene is then m = n * molar mass = 108.14 moles * 28 g/mol = 3027.92 g or approximately 3.03 kg of ethylene.
2. Na2SiO3(s) + HF(aq) !H2SiF6(aq) + NaF(aq) + H2O (l)
a. Balance the reaction.
b.How many moles of HF are needed to react with 0.300 mol of Na2SiO3?
c.How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?
d.How many grams of Na2SiO3can react with 0.800 g of HF?
e. Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?
Answer:
a) Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)
b) 2.40 moles HF
c) 5.25 grams NaF
d)0.609 grams Na2SiO3
e) 0.89 grams Na2SiO3
Explanation:
Step 1: Data given
Step 2: The balanced equation
Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)
b) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?
For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O
For 0.300 moles Na2SiO3 we'll need 8*0.300 = 2.40 moles HF
c. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?
For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O
For 0.500 moles we'll have 0.500 / 4 = 0.125 moles NaF
Mass NaF = 0.125 moles * 41.99 g/mol
Mass NaF = 5.25 grams NaF
d. How many grams of Na2SiO3 can react with 0.800 g of HF?
Moles HF = 0.800 grams / 20.01 g/mol
Moles HF = 0.0399 moles
For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O
For 8 moles we need 1 moles Na2SiO3 to react
For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3
Mass Na2SiO3 = 0.00499 moles * 122.06 g/mol
Mass Na2SiO3 = 0.609 grams Na2SiO3
Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?
Number of moles HF = 0.0399 moles
Number of moles Na2SiO3 = 1.5 grams / 122.06 g/mol = 0.0123 moles
For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O
For 8 moles we need 1 moles Na2SiO3 to react
For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3
There will remain 0.0123 - 0.00499 = 0.00731 moles Na2SiO3
Mass Na2SiO3 remaining = 0.00731 * 122.06 g/mol = 0.89 grams Na2SiO3
Final answer:
The balanced chemical equation requires 6 moles of HF for every mole of Na2SiO3. For different amounts of reactants, stoichiometry is used to calculate the moles and grams of other substances involved in the reaction.
Explanation:
The balanced chemical equation for the reaction between sodium metasilicate (Na2SiO3) and hydrofluoric acid (HF) is as follows:
Na2SiO3(s) + 6HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O(l)
b. According to the balanced equation, 6 moles of HF are needed to react with 1 mole of Na2SiO3. Therefore, to react with 0.300 mol of Na2SiO3, you would need 1.8 moles of HF (0.300 mol x 6).
c. Again according to the balanced equation, every 6 moles of HF produce 2 moles of NaF. So when 0.500 mol of HF reacts, it produces (0.500 mol HF x 2 mol NaF / 6 mol HF) = 0.167 moles of NaF. The molar mass of NaF is approximately 41.99 g/mol, so the mass of NaF formed is (0.167 mol x 41.99 g/mol) = 7.01 grams.
d. Given the molar mass of HF is approximately 20.01 g/mol, 0.800 g of HF corresponds to (0.800 g / 20.01 g/mol) = 0.0400 moles of HF. Using the balanced equation, we can determine the amount of Na2SiO3 that can react: (0.0400 mol HF x 1 mol Na2SiO3 / 6 mol HF) = 0.00667 moles of Na2SiO3. The molar mass of Na2SiO3 is approximately 122.06 g/mol, so the mass of Na2SiO3 that can react is (0.00667 mol x 122.06 g/mol) = 0.814 grams.
e. If you start with 1.5 g of Na2SiO3 and only 0.814 grams can react with the given 0.800 g of HF, then the amount of Na2SiO3 left after the reaction is (1.5 g - 0.814 g) = 0.686 grams.
After heating your beaker of Cu(OH)2, you formed copper (II) oxide. Before you added the H2SO4, you were instructed to decant the supernatant, add warm water, and decant again. What was the purpose of this step
Explanation:
The precipitate formed that is copper (II) oxide is heavy and coarse so the technique called Decantation is done to drain the supernatant liquid. The purpose of this step is to obtain a liquid that is free from particulates usually known as decanting. Decantation is a process in which a supernatant liquid is drained over by making the container somewhat tilt. In this method, before adding the sulphuric acid, the supernatant is filtered out and the precipitate is washed out.5. At 20°C, the water autoionization constant, Kw, is 6.8 ´ 10–15. What is the H3O+ concentration in neutral water at this temperature? A. 6.8 × 10–7 M B. 3.4 × 10–15 M C. 6.8 × 10–15 M D. 8.2 × 10–8 M E. 1.0 × 10–7 M
Explanation:
Let us assume that the concentration of [[tex]OH^{-}[/tex] and [tex]H^{+}[/tex] is equal to x. Then expression for [tex]K_{w}[/tex] for the given reaction is as follows.
[tex]K_{w} = [OH^{-}][H^{+}][/tex]
[tex]K_{w} = x^{2}[/tex]
[tex]6.8 \times 10^{-15} = x^{2}[/tex]
Now, we will take square root on both the sides as follows.
[tex]\sqrt{6.8 \times 10^{-15}} = \sqrt{x^{2}}[/tex]
[tex][H^{+}] = 8.2 \times 10^{-8}[/tex] M
Thus, we can conclude that the [tex]H_{3}O^{+}[/tex] concentration in neutral water at this temperature is [tex]8.2 \times 10^{-8}[/tex] M.
Answer: The concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]
Explanation:
The chemical equation for the ionization of water follows:
[tex]2H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The expression of [tex]K_w[/tex] for above equation, we get:
[tex]K_w=[H_3O^+]\times [OH^-][/tex]
We are given:
[tex]K_w=6.8\times 10^{-15}[/tex]
[tex][H^+]=[OH^-]=x[/tex]
Putting values in above equation, we get:
[tex]6.8\times 10^{-15}=x\times x\\\\x=8.2\times 10^{-8}M[/tex]
Hence, the concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]
A mixture of 0.577 M H_2O , 0.314 M Cl_2O , and 0.666 M HClO are enclosed in a vessel at 25°C .
H_2O(g) + Cl_2O(g) <-------> 2 HOCl (g) Kc = 0.0900 at 25°C
1. Calculate the equilibrium concentrations of each gas at 25°C .
Answer:
Equilibrium Concentration of H₂O(g) = 0.803
Equilibrium Concentration of Cl₂O(g) = 0.540
Equilibrium Concentration of HOCl (g) = 0.214
Explanation:
Given;
H₂O(g) + Cl₂O(g) <-------> 2HOCl (g)
I 0.577 0.314 0.666
C - x -x +2x
E 0.577 - x 0.314 - x 0.666 +2x
[tex]K_c = \frac{[HOCL]^2}{[H_2O][CL_2O]} \\\\0.09 = \frac{[0.666+2x]^2}{[0.577-x][0.314-x]}\\\\0.09(0.1812 -0.891x+x^2) = (0.666+2x)(0.666+2x)\\\\0.0163-0.0802x+0.09x^2 = 0.4436+2.664x+4x^2\\\\3.91x^2+2.7742x+0.4273 =0\\\\x = -0.226, or -0.483[/tex]
Equilibrium Concentration of H₂O(g) = 0.577 - (- 0.226) = 0.803
Equilibrium Concentration of Cl₂O(g) = 0.314 - (- 0.226) = 0.540
Equilibrium Concentration of HOCl (g) = 0.666 +2(- 0.226) = 0.214
Thus, from the result it can be seen that at equilibrium, the reactants are favored.
A chemist needs to know the mass of a sample of to significant digits. She puts the sample on a digital scale. This is what the scale shows: 0 0 7 6 . 2 1 g If this measurement is precise enough for the chemist, round it to significant digits. Otherwise, press the "No solution" button.
Final answer:
The mass 0076.21 g should be rounded to two significant figures as 76 g, omitting leading zeros, which are not significant. Significant figures indicate precision and uncertainty in scientific measurements.
Explanation:
The digital scale used by the chemist displays the mass of the sample as 0076.21 grams. To report the mass with two significant digits, we round the figure to 76 grams. This is because the leading zeros do not count towards significant figures; they are only placeholders to indicate the decimal position. The '7' and '6' are the two significant digits from the number 0076.21 grams.
It is crucial to use the correct number of significant figures in scientific measurements because they communicate the precision of the measurement and the uncertainty. For instance, if a balance measures to the nearest ±0.1 mg, one should not report a mass with more precision than this, as it would suggest a false level of accuracy. For addition and subtraction, the result must not have more decimal places than the measurement with the least precision. So if the scales used have different precisions, the final weight should be reported with the precision of the least precise scale.
Consider the reaction N2(g) + 3 H2(g) <=> 2 NH3(g), which is exothermic as written. What would be the effect on the equilibrium position of adding N2(g)?
Answer:
Shift to the right side
Explanation:
Remember that any stress that we bring to a chemical equilibrium, will be answered by this in such a way as to minimize the disturbance and attain equilibrium again. This is known as LeChatelier´s principle.
If we add N₂(g) to the system at equilibrium
N₂ ( g) + 3 H₂(g) ⇄ 2 NH₃(g)
the equilibrium will consume part of the added N₂ gas until it reaches equilibrium again producing more ammonia (NH₃).
Mathematically this is equivalent to a decrease the pressure of N₂ so that the equilibrium partial pressures of the gases obey the equilbrium pressure constant Kp.
Therefore the equilibrium will shift to the product side.
Final answer:
Adding N2(g) to the equilibrium system shifts the equilibrium towards the formation of more NH3(g), as the system seeks to counteract the increase in N2(g) concentration by consuming it, along with H2(g), to produce additional NH3(g).
Explanation:
The question asks about the effect of adding N2(g) to the equilibrium system of the reaction N2(g) + 3 H2(g) ⇆ 2 NH3(g), which is exothermic. According to Le Chatelier's Principle, if a system at equilibrium is disturbed by a change in concentration, temperature, or pressure, the system will adjust itself to counteract that disturbance and a new equilibrium will be established. In this case, adding more N2(g) shifts the equilibrium towards the products to reduce the disturbance, thereby increasing the concentration of NH3(g) and using up some of the added N2(g) along with H2(g) to reestablish equilibrium. This movement towards the products is a direct response to the increased concentration of a reactant, demonstrating the system's attempt to maintain equilibrium.
For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?You do not need to look up any values to answer this question. Check all that apply. View Available Hint(s) Check all that apply. CO(g)+12O2(g)→CO2(g) Li(s)+12F2(l)→LiF(s) C(s)+O2(g)→CO2(g) CaCO3(g)→CaO+CO2(g) 2Li(s)+F2(g)→2LiF(s) Li(s)+12F2(g)→LiF(
Final answer:
ΔH∘rxn is equal to ΔH∘f of the product(s) for the reactions CO(g) + 1/2O2(g) → CO2(g) and 2Li(s) + F2(g) → 2LiF(s).
Explanation:
In chemistry, the enthalpy change of a reaction (ΔH°rxn) is equivalent to the enthalpy of formation (ΔH°f) of the product(s) when the reaction involves the formation of a compound from its elements in their standard states.
The reactions for which ΔH∘rxn is equal to ΔH∘f of the product(s) are:
CO(g) + 1/2O2(g) → CO2(g)2Li(s) + F2(g) → 2LiF(s)For these reactions, the enthalpy change of the reaction is equal to the standard enthalpy of formation of the product(s). The enthalpy change of the reaction can be calculated using Hess's law and the enthalpy changes of the individual reactions.
Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows: C5H5N+H2O⇌C5H5NH++OH− The pKb of pyridine is 8.75. What is the pH of a 0.460 M solution of pyridine? (Assume that the temperature is 25 ∘C.) Express the pH numerically to two decimal places.
Answer:
The pH of the solution is 9.46
Explanation:
C₅H₅N + H₂O ⇌ C₅H₅NH+OH⁻
I 0.46
C - x +x +x
E 0.46 - x
-LogKb = Pkb
[tex]Kb =10^{[-PKb]} = 10^{[-8.75]} = 1.778 X 10^{-9}[/tex]
[tex]Kb = \frac{[C5H5NH][OH^-]}{[C5H5N]}[/tex]
[tex]1.778 X10^{-9}= \frac{X^2}{0.46-X} \\\\X^2 = 8.1788 X 10^{-10} - 1.778 X10^{-9}X\\\\X^2 + 1.778 X10^{-9}X -8.1788 X 10^{-10}\\\\X = 2.85977 X 10^{-5} = [OH^-][/tex]
pOH = -Log[OH⁻]
pOH = -Log [2.85977 x 10⁻⁵]
pOH = 4.54
pOH + pH = 14
pH = 14 - pOH
pH = 14 - 4.54
pH = 9.46
Therefore, the pH = 9.46
what is the density of a substance that has a mass of 2.0 g, and when placed in a graduated cylinder the volume rose from 70 mL to 75 mL
Answer:
0.4g/mL
Explanation:
The following data were obtained from the question:
Mass = 2g
Volume = 75 — 70 = 5mL
Density =?
Density = Mass /volume
Density = 2/5
Density = 0.4g/mL
The density of the substance is 0.4g/mL
The density of a substance that has a mass of 2.0 g, and when placed in a graduated cylinder the volume rose from 70 mL to 75 mL is 0.4 g/mL.
To find the density of the substance, we need to calculate the volume it displaces and then divide its mass by that volume.
The volume displaced by the substance can be found by subtracting the initial volume in the graduated cylinder from the final volume after the substance is added. The initial volume is 70 mL and the final volume is 75 mL. Therefore, the volume displaced by the substance is:
Volume displaced = Final volume - Initial volume
Volume displaced = 75 mL - 70 mL
Volume displaced = 5 mL
Now, we have the mass of the substance, which is 2.0 g, and the volume it displaces, which is 5 mL. The density [tex]\rho[/tex] is calculated by dividing the mass (m) by the volume (V):
[tex]\rho = \frac{m}{V}[/tex]
[tex]\rho = \frac{2.0 \text{ g}}{5 \text{ mL}}[/tex]
[tex]\rho = 0.4 g/mL[/tex]
Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 , K a2 = 6.2 × 10 − 8 , and K a3 = 4.8 × 10 − 13 ). To find the pH of a buffer composed of H 2 PO − 4 ( aq ) and HPO 2 − 4 ( aq ) , which p K a value should be used in the Henderson–Hasselbalch equation?
To calculate the pH of a buffer solution composed of H2PO−4(aq) and HPO2−4(aq), the specific pKa value for the ionization step of interest needs to be known.
Explanation:The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution.
In the case of phosphoric acid, which is a triprotic acid, we need to use the pKa value that corresponds to the ionization step we are interested in.
Since the question does not specify which buffer composition is being used, we cannot determine which pKa value to use from the information given.
A constant volume calorimeter (bomb calorimeter) was calibrated by performing in it a reaction in which 5.23 kJ of heat energy was released, causing the calorimeter to rise by 7.33 °C. What is the heat capacity C of the calorimeter?
Explanation:
Relation between heat energy and specific heat is as follows.
[tex]q_{bomb} = c \times \Delta t[/tex]
or, c = [tex]\frac{q_{bomb}}{\Delta t}[/tex]
where, c = specific heat
[tex]q_{bomb}[/tex] = heat energy
[tex]\Delta t[/tex] = change in temperature
Putting the given values into the above formula we will calculate the specific heat as follows.
c = [tex]\frac{q_{bomb}}{\Delta t}[/tex]
= [tex]\frac{5.23 kJ}{7.33^{o}C}[/tex]
= 0.713 [tex]kJ^{o}C[/tex]
Thus, we can conclude that heat capacity C of the calorimeter is 0.713 [tex]kJ^{o}C[/tex].
Final answer:
The heat capacity C of the calorimeter is calculated using the formula C = q / ΔT, with the given values resulting in a C of 0.7138 kJ/°C.
Explanation:
The heat capacity C of the calorimeter can be calculated using the amount of heat energy released and the resultant temperature change. The formula to calculate heat capacity is C = q / ΔT, where q is the heat energy released and ΔT is the change in temperature.
In this case, 5.23 kJ of heat energy was released, causing a temperature rise of 7.33 °C. Therefore, the heat capacity of the calorimeter is calculated as follows:
C = 5.23 kJ / 7.33 °C = 0.7138 kJ/°C
It's crucial to note that the heat capacity is commonly expressed in kJ/°C for bomb calorimeters, reflecting the amount of energy required to raise the temperature of the entire calorimeter setup by one degree Celsius.
Choose the answer in which the three atoms and/or ions are listed in order of increasing EXPECTED size (smallest particle listed first) assuming that the simple shell model of the atom is correct. (Neglect quantum, relativistic and other advanced considerations.)
a) Cl - < S2- < Ar
b) Ar < Cl - < S2-
c) Ar < S2- < Cl -
d) Cl - < Ar < S2-
e) S2- < Cl - < Ar
Answer:
Ar < Cl - < S2-
Explanation:
All the species written above are isoelectronic. This means that they all possess the same number of electrons. All the species above possess 18 electrons, the noble gas electron configuration.
However, for isoelectronic species, the greater the atomic number of the specie, the smaller it is. This is because, greater atomic number implies that their are more protons in the nucleus exerting a greater attractive force on the electrons thereby making the specie smaller in size due to high electrostatic attraction.
The correct order of increasing expected size among Cl-, S2-, and Ar based on the simple shell model and their respective nuclear charges—considering they are all isoelectronic—is Ar < Cl- < S2-.
Explanation:The question asks to arrange the atoms and/or ions Cl-, S2-, and Ar in order of increasing expected size based on the simple shell model of atomic radius. In the simple shell model, atomic radius increases with the addition of electron shells and decreases with an increase in nuclear charge for isoelectronic species, meaning species with the same number of electrons. Cl-, S2-, and Ar are all isoelectronic with the closed-shell electron configuration of [Ar]. The Cl- ion has 17 protons, S2- has 16 protons, and Ar has 18 protons.
Following this reasoning:
Cl-, with 17 protons, is smaller than S2-, which has fewer protons (16) and thus less nuclear charge to pull on the same number of electrons.Ar, with 18 protons, is even smaller than Cl- because it has a larger nuclear charge.Therefore, the correct order from smallest to largest is: Ar < Cl- < S2-.
Write the net cell equation for this electrochemical cell. Phases are optional. Do not include the concentrations. Co ( s ) ∣ ∣ Co 2 + ( aq , 0.0155 M ) ∥ ∥ Ag + ( aq , 3.50 M ) ∣ ∣ Ag ( s )
Answer: The net cell equation for the given electrochemical cell is given below.
Explanation:
The given chemical cell follows:
[tex]Co(s)|Co^{2+}(aq.,0.0155M)||Ag^{+}(aq,3.50M)|Ag(s)[/tex]
Oxidation half reaction: [tex]Co(s)\rightarrow Co^{2+}(aq,0.0155M)+2e^-[/tex]
Reduction half reaction: [tex]Ag^{+}(aq,3.50M)+e^-\rightarrow Ag(s)[/tex] ( × 2)
Net cell reaction: [tex]Co+2Ag^{+}(aq,3.50M)\rightarrow Co^{2+}(aq,0.0155M)+2Ag(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
Hence, the net cell equation for the given electrochemical cell is given above.
How many grams of CO2 are dissolved in a 1.00 L bottle of carbonated water at 298 K if the pressure used in the carbonation process was 1.8 bar? The density of water at this temperature is 998 kg⋅m−3. The Henry's law constant for aqueous solution of CO2 at this temperature is 1.65×103bar.
Using Henry's Law, the solubility of CO2 at 1.8 bar and 298 K is found to be 2970 g/m3. Since there is 1 L of water, this equates to 2.97 g of CO2 dissolved in the 1.00 L bottle of carbonated water.
Explanation:The question asks how many grams of CO2 are dissolved in a 1.00 L bottle of carbonated water at a pressure of 1.8 bar and 298 K using the Henry's Law constant for CO2. To find the concentration of CO2 in the water, we use Henry's Law which states that the solubility of a gas in a liquid is proportional to the pressure of that gas above the liquid. The Henry's Law equation is Sg = kH * Pg, where Sg is the solubility of the gas, kH is the Henry's Law constant and Pg is the partial pressure of the gas.
The density of water is given as 998 kg/m3, which we use to convert the volume of water to mass. Since 1 L of water has a mass of approximately 998 g and the Henry's Law constant for CO2 at 298 K is 1.65×103 bar, we can plug in our values: Sg = (1.65×103 bar)(1.8 bar) = 2970 g/m3 of CO2. Since we have 1 L of water, or 0.998 kg, we can convert to grams: 0.998 kg × 1000 g/kg = 998 g.
To find the grams of CO2 dissolved, we use the solubility in g/m3: 2970 g/m3 × 0.001 m3 = 2.97 g of CO2.
Draw the structure of a compound with the molecular formula C9H10O2 that exhibits the following spectral data. (a) IR: 3005 cm-1, 1676 cm-1, 1603 cm-1 (b) 1H NMR: 2.6 ppm (singlet, I
Answer:
The answer is 3-Phenylpropanoic acid (see attached structure)
Explanation:
From spectral data:
3005 cm-1 ⇒ carboxylic acid (broad band)
1670 cm-1 ⇒ C=C
1603 cm-1 ⇒ Aromatic C-C bond
H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding (clearer investigation of spectrum would be expedient).
Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.
The compound is probably a molecule with an aromatic ring (benzene) with an attached ester group (COOCH3) that satisfies all the given spectral data.
Explanation:The spectral data given corresponds to a compound with molecular formula C9H10O2. From the IR data, the bands at 3005 cm-1 suggests C-H sp2 bond (alkene or aromatic), at 1676 cm-1 indicates carbonyl group (C=O), and 1603 cm-1 suggests a carbon-carbon double bond (C=C) which might be in an aromatic ring. The 1H NMR data: 2.6 ppm singlet signifies the protons of a methyl group (CH3) attached to an electronegative atom like a carbonyl carbon.
Based on these data, a probable structure for the compound could be a molecule, which is an aromatic ring (benzene) with an attached ester group (COOCH3). That gives the right molecular formula, the required carbonyl, alkene and sp2 hybridized C-H bonds for the IR, and the singlet in the NMR for the methyl group (CH3) of the ester.
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A student fills her burette with NaOH to the 2.5 mL mark. She titrated her sample of the NaOH until she reaches the endpoint (i.e. all the acid has been neutralized by the NaOH). The volume marking on her burette - at the endpoint - is 52.5 mL. How many mL of NaOH did she use to reach the endpoint
Answer:
She used 50 mL of NaOH to reach the endpoint.
Explanation:
Assuming the burette is filled to the point marked 3.30 ml, You would record the initial point as 3.30 ml:
If at the end of the titration the level of the NaOH is at 20.30 mL; Subtract the initial reading from the final burette reading to get how many mL was used to reach an end point.
That is 20.3 - 3.3 = 17.00 mL
Therefore, the titration would have required 17.00 mL.
Remember that you should read the number that is at the bottom of the meniscus and at an eye level in order to avoid error.
Initial mark = 2.5 mL
Final mark = 52.5 mL.
volume used = 52.5 - 2.5
= 50 mL
Suppose you want to separate a mixture of the following compounds: salicylic acid, 4-ethylphenol, p-aminoacetophenone, and napthalene. Come up with a list of steps and chemicals needed to most efficiently isolate all four compounds as solids with the greatest purity possible. You do not need to write a formal procedure, but be sure to indicate steps needed clearly and in order.
Answer:
The procedure you will use in this exercise exploits the difference in acidity and solubility just described.
(a) you will dissolve your unknown in ethyl acetate (an organic solvent). All of the possible compounds are soluble in ethyl acetate.
(b) you will extract with sodium bicarbonate to remove any carboxylic acid that is present.
(c) you will extract with sodium hydroxide to remove any phenol that is present.
(d) you will acidify both of the resulting aqueous solutions to cause any compounds that were extracted to precipitate.
Use a spreadsheet and construct curves for the following titrations. Calculate potentials after the addition of 10.00, 25.00, 49.00, 49.90, 50.00, 50.10, 51.00, and 60.00 mL of the reagent. Where necessary, assume that = 1.00 throughout.