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4 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?

1 answer:

Usimov [2.4K]4 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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3 years ago

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Answer:

<u>It increases by a factor of four</u>

Explanation:

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Applying equation (1) and (2)

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According to question ,

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3 years ago

What is △n for the following equation in relating Kc to Kp?

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Explanation:

The relation between Kp and Kc is given below:

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4 years ago

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24.6 ℃

<h3>Explanation</h3>

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which is equivalent to

$\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)$

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Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

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3 years ago