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Hunter-Best [27]
2 years ago
8

(a)Determine the number of KNO3 molecules in 0.750 mol KNO3.

Chemistry
1 answer:
svp [43]2 years ago
5 0

A. The number of molecules in 0.750 mole of KNO₃ is 4.515×10²³ molecules

B. The mass (in milligrams) of 2.39×10²⁰ molecules of Ag₂SO₄ is 124 mg

C. The number of molecules in 3.429 g of NaHCO₂ is 3.04×10²² molecules

<h3>Avogadro's hypothesis </h3>

1 mole of substance = 6.02×10²³ molecules

<h3>A. How to determine the number of molecules </h3>

1 mole of KNO₃ = 6.02×10²³ molecules

Therefore,

0.750 mole of KNO₃ = 0.75 × 6.02×10²³

0.750 mole of KNO₃ = 4.515×10²³ molecules

<h3>B. How to determine the mass of Ag₂SO₄</h3>

6.02×10²³ molecules = 312 g of Ag₂SO₄

Therefore,

2.39×10²⁰ molecules = (2.39×10²⁰ × 312) / 6.02×10²³

2.39×10²⁰ molecules = 0.124 g

Multiply by 1000 to express in mg

2.39×10²⁰ molecules = 0.124 g × 1000

2.39×10²⁰ molecules = 124 mg of Ag₂SO₄

<h3>C. How to determine the number of molecules </h3>

68 g of NaHCO₂ = 6.02×10²³ molecules

Therefore,

3.429 g of NaHCO₂ = (3.429 × 6.02×10²³) / 68

3.429 g of NaHCO₂ = 3.04×10²² molecules

Learn more about Avogadro's number:

brainly.com/question/26141731

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You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f
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Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

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V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65  K

Putting values in above equation, we get:

(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole

Percentage recovery of carbon dioxide gas =  49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

\frac{49.4}{100}\times  0.0438 mol=0.02164 mol

2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

\frac{2}{1}\times 0.02164 mol=0.04328 mol

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

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A warning label is required to provide set information. The following list contains only some of this required information. What
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If a 2.50 liter container is filled with Ne gas at a pressure of 650 mm Hg and at 25°C, what mass of Ne is in the container?
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Answer:

1.76 g is the mass of Ne is in the container.

Explanation:

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 650 mm Hg

V = Volume of the gas = 2.50 L

T = Temperature of the gas = 25^oC=[25+273]K=298K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

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Putting values in above equation, we get:

650mmHg\times 2.50L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{650\times 2.50}{62.3637\times 298}=0.0874mol

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