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Bingel [31]
3 years ago
11

The concentration of hydrogen peroxide (H2O2) can be determined by titrating it with an acidified MnO4− solution. The following

is an unbalanced equation.
MnO4− (aq) + H2O2 (aq) → O2 (g) + Mn2+ (aq)

(a) Balance the above redox reaction.
(b) Determine the concentration of H2O2 solution in molarity if 10.00 mL of this solution requires 20.00 mL of 1.5 M MnO4− for a complete reaction.
Chemistry
1 answer:
Vaselesa [24]3 years ago
5 0

Answer:

A) {MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2

B) 7.5 molar

Explanation:

A) <u>Reduction</u>

{MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O

<u>Oxidation</u>

H_2O_2 \rightarrow O_2 + 2 H^+ + 2e^-

Multiplying the oxidation reaction by 5/2 and adding it to the reduction equation:

{MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O

+

\frac{5}{2} H_2O_2 \rightarrow \frac{5}{2} O_2 + 5 H^+ + 5e^-

-----------------------------------------------------------------------------------------------------

{MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2

B) 10 ml = 0.01 L

20 ml = 0.02 L

mol of MnO4− = molarity*volume = 1.5*0.02 = 0.03

1 mol of MnO4− reacts with 5/2 mol of H2O2, then:

mol of H2O2 = 0.03*5/2 = 0.075

molarity = mol/volume = 0.075/0.01 = 7.5 molar

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The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0
4 years ago
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
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Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

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We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

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To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

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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 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

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