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agasfer [191]
3 years ago
15

A student wants to make a 0.150 M aqueous solution of silver (I) nitrate but only has 11.27 g of AgNO3. What volume of the 0.150

M solution can they make?
Chemistry
1 answer:
Usimov [2.4K]3 years ago
3 0

Answer:

442.3 mL

Explanation:

Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:

M=\frac{Moles of substance X}{Volume of the solution}

Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:

11.27 g AgNO_{3} *\frac{1 mole AgNO_{3}}{169.87 g AgNO_{3}} = 0.06634 moles AgNO_{3}

Then you can solve for the volume of the solution:

Volume of the solution=\frac{Moles of AgNO_{3}}{M} =\frac{0.06634 mol AgNO_{3}}{0.150 M} =0.4423 L = 442.3 mL

Hope it helps!

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The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process
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0.1066 hours

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7 0
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Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe
jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

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