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3 years ago

The value of ΔGo for the precipitation reaction,

1 answer:

5 0

Answer:

$\Delta G=-44.9kJ$

Explanation:

We know, $\Delta G=\Delta G^{0}+RTlnQ$

where Q is the reaction quotient and for this reaction, it is expressed as-

$Q=\frac{1}{[Ca^{2+}][CO_{3}^{2-}]}$

where species under third bracket represent concentrations in molarity.

T represents temperature in kelvin scale

Here T = 298 K

R = gas constant = 8.314 J/K

So, $\Delta G=(-48.1\times 10^{3}J)+(8.314\frac{J}{K}\times 298K)ln(\frac{1}{0.312\times 0.898})$

or, $\Delta G=-44948J$

or, $\Delta G=-44.9kJ$

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stira [4]

Answer:

%KCl = 7.05%

%Water = 92.95%

Explanation:

Step 1: Given data

Step 2: Calculate the mass of the solution

The mass of the solution is equal to the sum of the masses of the solute and the solvent.

m = 36 g + 475 g = 511 g

Step 3: Calculate the mass percentage of the solution

We will use the following expression.

%Component = mComponent/mSolution × 100%

%KCl = 36 g/511 g × 100% = 7.05%

%Water = 475 g/511 g × 100% = 92.95%

7 0

3 years ago

scoray [572]

Answer:

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

Explanation:

Oxidation: $As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)$

Balance As: $As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)$
Balance H and O in acidic medium: $As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$
Balnce charge: $As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$ ......(1)

Reduction: $NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$

Balance N: $2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$
Balance H and O in acidic medium: $2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$
Balance charge: $2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$ ......(2)