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jasenka [17]
3 years ago
12

If 97.7 ml of silver nitrate solution reacts with excess potassium chloride solution to yield 0.326 g of precipitate, what is th

e molarity of silver ion in the original solution?
Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
4 0

The complete balanced chemical reaction is written as:

AgNO3 + KCl ---> AgCl + KNO3 

where AgCl is our precipitate

 

So calculating for moles of AgCl produced: MM AgCl = 143.5 g/mol

moles AgCl = 0.326 g / (143.5 g/mol) = 2.27 x 10^-3 mol

we see that there is 1 mole of Ag per 1 mole of AgCl so:

moles Ag = 2.27 x 10^-3 mol

 

The molarity is simply the ratio of number of moles over volume in Liters, therefore:

Molarity = 2.27 x 10^-3 mol / 0.0977 L

<span>Molarity = 0.0233 M</span>

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Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

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harkovskaia [24]

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Explanation:

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Magnesium has a density of 1.738 g/cmº at 25°C. What is the mass of a block of
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3 years ago
[25 pts] When a 14.2g sample of a compound containing only mercury and oxygen is decomposed, 13.2g of Hg is obtained. What is th
Agata [3.3K]

Answer:

% composition of Hg = 93 %

% composition of O = 7 %

Explanation:

Data Given:

mass of total sample = 14.2 g

mass of mercury (Hg) = 13.2 g

percent composition of Hg =

percent composition of O =

Solution:

First find mass of oxygen

As this compound only have two components that is oxygen and mercury

So

upon decomposition it gives 13.2 g mercury

now to find mass of oxygen we have to subtract mass of mercury from total mass of sample

                 Mass of oxygen =  Total mass of sample - mass of Mercury

                 Mass of oxygen =  14.2 g - 13.2 g

                 Mass of oxygen =  1 g

Now to find percent composition of Hg and Oxygen

% composition = component mass / total mass of sample x 100. . . .(1)

For % composition of mercury Hg

Put values in equation 1

             % composition of Hg = 13.2 g / 14.2 g x 100

               % composition of Hg = 93%

For % composition Oxygen (O)

Put values in equation 1

             % composition of O = 1 g / 14.2 g x 100

               % composition of O = 7 %

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