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natali 33 [55]
3 years ago
14

Electrochemical cells can be used to measure ionic concentrations. A cell is set up with a pair of Zn electrodes, each immersed

in ZnSO4 solution. One solution is created to be exactly 0.300 M, and the potential on its electrode is measured to be 0.045 V higher than the other electrode. What is the concentration of the other solution
Chemistry
1 answer:
Phantasy [73]3 years ago
6 0

Answer:

the concentration of the solution is 0.00906 M

Explanation:

Given the data in the question;

we know that from Nernst Equation;

E = E⁰ - ((0.0592/n) logQ)

now, E₀ for concentration cell is 0

n for this redox is 2

concentration of the unknown solution is x

so we substitute

0.045 = 0 - ( 0.0592 / 2)log( x/0.300 ))

0.045 = -0.0296log( x/0.300 )

divide both side by 0.0296

1.52 = -log( x/0.300 )

x/0.300 = 10^{-1.52

x/0.300 = 0.0301995

we cross multiply

x = 0.300 × 0.0301995

x =  0.00906 M

Therefore, the concentration of the solution is 0.00906 M

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Nitric oxide (NO) reacts readily with chlorine gas as follows.2 NO(g) + Cl2(g) equilibrium reaction arrow 2 NOCl(g)At 700. K the
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<u>Answer:</u>

<u>For a:</u> The mixture will need to produce more reactants to reach equilibrium.

<u>For b:</u> The mixture will need to produce more reactants to reach equilibrium.

<u>For c:</u> The mixture will need to produce more products to reach equilibrium.

<u>Explanation</u>:

For the given chemical equation:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

The expression of K_{p} for above equation follows:

K_{p}=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}   .....(1)

We are given:

Value of K_p = 0.26

There are 3 conditions:

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  • When K_{p}; the reaction is reactant favored.
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For the given options:

  • <u>For a:</u>

We are given:

p_{NOCl}=0.11atm\\p_{NO}=0.16atm\\p_{Cl_2}=0.30atm

Putting values in expression 1, we get:

Q_p=\frac{(0.11)^2}{(0.16)^2\times 0.30}=1.57

As, K_{p}; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

  • <u>For b:</u>

We are given:

p_{NOCl}=0.048atm\\p_{NO}=0.12atm\\p_{Cl_2}=0.10atm

Putting values in expression 1, we get:

Q_p=\frac{(0.048)^2}{(0.12)^2\times 0.10}=1.6

As, K_{p}; the reaction is reactant favored

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  • <u>For c:</u>

We are given:

p_{NOCl}=5.20\times 10^{-3}atm\\p_{NO}=0.15atm\\p_{Cl_2}=0.15atm

Putting values in expression 1, we get:

Q_p=\frac{(5.20\times 10^{-3})^2}{(0.15)^2\times 0.15}=0.008

As, K_{p}>Q_p; the reaction is product favored.

Hence, the mixture will need to produce more products to reach equilibrium.

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How much heat is required to raise 36 g ice at – 10.0oC to steam at 110oC? (get your answers from question #1)
Gnom [1K]

Answer:

The total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

Explanation:

The heat involved in this process involves the following:

1. Heat to change ice at -10°C to ice at 0°C;

2. Heat to change ice at 0°C to water at 0°C

3. Heat to change water at 0°C to water at 100°C

4. Heat to change water at 100°C to steam at 100°C

5. Heat to change steam at 100°C to steam at 110°C

Specific heat capacity of ice, c = 2040 J/K/kg, Latent heat of fusion of ice, L = 3.35 × 10⁵ J/kg, specific heat capacity of water, c =  4182 J/K/kg, latent heat of vaporization of water, l = 2.26 × 10⁶ J/kg, specific heat capacity of steam, c = 1996 J/K/kg

Step 1: H = mcθ; where m = 30.0 g = 0.03 g, c = 2040 J/K/kg, θ = (0 -  -10) = 10 K

H = 0.03 * 2040 * 10 = 612 J

Step 2: H = mL, where  L = 3.35 × 10⁵ J/kg

H = 0.03 * 3.35 × 10⁵ = 10050 J

Step 3: H = mcθ, where c =  4182 J/K/kg, θ = (100 - 0) = 100 K

H = 0.03 * 4182 * 100 = 12546 J

Step 4: H = ml, where l = 2.26 × 10⁶

H = 0.03 * 2.26 × 10⁶ = 67800 J

Step 5: H = mcθ, where c = 1996 J/K/kg, θ = (110 - 100) = 10 K

H = 0.03 * 1996 * 10 = 598.8 J

Total heat required to raise ice at -10°C to steam at 110°C = (612 + 10050 + 12546 + 67800 + 598.8) J = 91606.8 J

Therefore, the total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

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