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PtichkaEL [24]
3 years ago
9

How much does the mass of 12.65-g sample of copper(II) nitrate hexahydrate decrease when heated?

Chemistry
1 answer:
garik1379 [7]3 years ago
4 0
There is 103 percent of water in hydrate
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A = ε l c. (a) Define the terms in the formula: A = ε l c. (Pick your answers using the letter of the correct definition.)
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Explanation:

Using Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

where,

A = absorbance of solution

c = concentration of solution

l = length of the cell

\epsilon = molar absorptivity of this solution

According to question:

A = (C) : absorbance measured by the spectrometer

c = (B) : concentration, in mol/L, of the stock solution from which the sample was made

l = (A): pathlength of light through the cell

ε =  (D) : molar absorptivity, a constant unique to that substance at that wavelength

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Which phase has the greatest structure
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4 years ago
Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl
Archy [21]

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = V

V_2 = final volume of gas = 2V

n_1 = initial moles of gas  = 1.56 mole

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

\frac{V}{2V}=\frac{1.56mole}{n_2}

n_2=3.12mole

Now we have to calculate the mass of helium gas at doubled volume.

\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He

\text{Mass of }He=3.12mole\times 4g/mole=12.48g

Therefore, the mass of helium gas added must be 12.48 grams.

4 0
3 years ago
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