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MariettaO [177]
3 years ago
12

Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su

re your answer has the correct number of significant digits.
Chemistry
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{175\ g}{101.543\ g/mol}

Moles_{copper(II)\ fluoride}= 1.7234\ mol

Also,

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

So,,

Volume =\frac{Moles\ of\ solute}{Molarity}

Given, Molarity = 0.00231 M

So,

Volume =\frac{1.7234}{0.00231}\ L

<u>Volume = 746 L</u>

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2 years ago
a sample of helium occupies a volume of 101.2 mL at a pressure of 790 mmHg. at what pressure would the volume be 120 mL?
AveGali [126]

Answer : The final pressure will be, 666.2 mmHg

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 790 mmHg

P_2 = final pressure = ?

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Now put all the given values in the above equation, we get:

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6 0
3 years ago
Calculate the volume in liters of a potassium iodide solution that contains of potassium iodide . Be sure your answer has the co
Anna11 [10]

Answer:

23.8 L

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.</em>

<em />

The molar mass of potassium iodide is 166.00 g/mol. The moles corresponding to 150 grams are:

150 g × (1 mol/166.00 g) = 0.904 mol

0.904 moles of potassium iodide are contained in an unknown volume of a 0.0380 mol/L potassium iodide solution. The volume is:

0.904 mol × (1 L/0.0380 mol) = 23.8 L

6 0
3 years ago
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