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3 years ago

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You have a solid 7.00 gram mixture of sodium nitrate and silver nitrate. You add distilled water to dissolve the solids. Now you

have aqueous solutions of sodium nitrate and silver nitrate. Next you add excess sodium chloride which results in a precipitate forming. You collect and dry the precipitate that forms and it has a mass of 2.54 grams. Write a balanced net ionic equation for the reaction that occurred. Determine the percent silver nitrate in the original mixture by mass assuming 95.9% actual yield.

1 answer:

olganol [36]3 years ago

3 0

Answer:

Net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

44.9% as AgNO₃

Explanation:

When sodium nitrate, NaNO₃ and silver nitrate, AgNO₃ are dissolved in water, the Na⁺, NO₃⁻ and Ag⁺ ions are formed.

Then, the addition of NaCl (Na⁺ and Cl⁻) produce AgCl⁻ as precipitate. <em>The net ionic equation is:</em>

<h3>Ag⁺(aq) + Cl⁻(aq) → AgCl(s)</h3><h3 />

If 2.54g of AgCl are formed and represents the 95.9% of yield. The real amount of AgCl is:

2.54g AgCl * (100% / 95.9%) = 2.65g AgCl.

In moles (Molar mass AgCl = 143.32g/mol):

2.65g AgCl * (1mol / 143.32g) = 0.0185 moles of AgCl = Moles of AgNO₃

<em>Because all Ag comes from AgNO₃</em>

<em />

Thus, the original mass of silver nitrate and its precentage is (Molar mass AgNO₃ = 169.87g/mol):

0.0185 moles AgNO₃ * (169.87g / mol) = 3.14g of AgNO₃

Percentage:

3.14g AgNO₃ / 7.00g * 100 =

<h3>44.9% as AgNO₃</h3>

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Calculate the ph at the equivalence point for the titration of a solution containing 150.0 mg of ethylamine (c2h5nh2) with 0.100

barxatty [35]

Answer:

pH = 6.2.

Explanation:

• Firstly, we need to calculate the no. of moles (n) of ethylamine (C₂H₅NH₂) using the law:

n = mass / molar mass,

Mass of C₂H₅NH₂ = 150.0 mg = 0.150 g.

Molar mass of C₂H₅NH₂ = 45.0847 g/mol.

∴ The no. of moles (n) of C₂H₅NH₂ = mass / molar mass = (0.150 g) / (45.0847 g/mol) = 0.00333 mol.

The ionic equation of the equivalence of ethylamine (C₂H₅NH₂) with HCl:

CH₃CH₂NH₂ + H⁺ ↔ CH₃CH₂NH₃⁺

The no of moles of CH₃CH₂NH₃⁺ = 0.00333 mol.

The molarity of (CH₃CH₂NH₃⁺) can be calculated by dividing its no. of moles (0.00333 mol) by the volume of the solution (0.250 L).

[CH₃CH₂NH₃⁺] = 0.00333 mol/ 0.250 L = 0.0133 M.

There is an equilibrium between the resulting (CH₃CH₂NH₃⁺) and water:

CH₃CH₂NH₃⁺ + H₂O ↔ CH₃CH₂NH₂ + H₃O⁺.

The CH₃CH₂NH₃⁺ decomposed by an amount x and (CH₃CH₂NH₂ & H₃O⁺) formed by amount x.

The hydrolysis constant Ka = Kw / Kb = (1.0 x 10⁻¹⁴) / (4.7 x 10⁻⁴) = 2.1 x 10⁻¹¹.

At equilibrium:

[CH₃CH₂NH₃⁺] = 0.0133 - x

[H₃O⁺] = [CH₃CH₂NH₂] = x

Ka = [CH₃CH₂NH₂] [H₃O⁺] / [CH₃CH₂NH₃⁺] = (x)(x) / (0.0133 -x)

2.1 x 10⁻¹¹ = (x)(x)/ 0.0133-x

x = [H₃O⁺] = 5.32 x 10⁻⁷ mol/L.

Also, we cannot neglect the [H₃O⁺] from the water dissociation

2H₂O ↔ H₃O⁺ + OH⁻

Kw = 1.0 x 10⁻¹⁴ = [H₃O⁺][OH⁻]

[H₃O⁺] = 1.0 x 10⁻⁷ mol/L.

The total concentration of (H₃O⁺) = 5.32 x 10⁻⁷ + 1.0 x 10⁻⁷ = 6.32 x 10⁻⁷ mol/L.

pH = - log [H₃O⁺] = - log (6.32 x 10⁻⁷)

pH = 6.20 .

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3 years ago

in the laboratory a student dilute 13.3 ml of a 10.8 m hcl solution to a total of 300.0 ml what is the concentration of the dilu

krek1111 [17]

<u>Answer:</u> The concentration of the diluted solution is 0.4788M

<u>Explanation:</u>

To calculate the concentration of the diluted solution, we use the following equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of HCl solution

$M_2\text{ and }V_2$ are the molarity and volume of the diluted HCl solution.

We are given:

$M_1=10.8M\\V_1=13.3mL\\M_2=?M\\V_2=300mL$

Putting values in above equation, we get:

$10.8\times 13.3=M_2\times 300\\\\M_1=0.4788M$

Hence, the concentration of the diluted solution is 0.4788M

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3 years ago

Point E is located at (-2, 2) and point F is located at (4-6). What is the distance between

Flauer [41]

Answer:

10

Explanation:

Given endpoints:

E (-2,2) and F(4, -6)

To find the distance between two points;

Use the expression below;

D = $\sqrt{(y_{2} - y_{1} )^{2} + (x_{2} - x_{1})^{2} }$

x₁ = -2,

x₂ = 4

y₁ = 2

y₂ = -6

Insert the parameters and solve;

D = $\sqrt{(-6 - 2 )^{2} + (4 -+ 2)^{2} }$

D = $\sqrt{64+ 36 }$ = 10

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4 years ago