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2 years ago

How many moles are in 428 L of chlorine gas

Chemistry

1 answer:

djyliett [7]2 years ago

6 0

Answer:

19.3 moles

Explanation:

In order to calculate the number of moles of Chlorine gas from the volume, we have to calculate its mass from its known density using the formula:

Density (g/L) = mass (g) /volume (L)

Density of Chlorine gas= 3.2g/L

Volume = 428L

3.2 = mass/428

mass = 428 × 3.2

mass = 1369.6g

Using mole = mass ÷ molar mass

Molar mass of chlorine (Cl2) = 35.5 × 2

= 71g/mol

mole = 1369.6 ÷ 71

mole = 19.2901

mole = 19.3 moles

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A sample of a gas has a volume of 852 mL at 298 K. If the gas is cooled to 200K, what would the new volume be?

Colt1911 [192]

Answer:

571.81 mL

Explanation:

Assuming constant pressure, we can solve this problem by using Charles' law, which states that at constant pressure:

V₁T₂=V₂T₁

Where in this case:

V₁ = 852 mL

T₂ = 200 K

V₂ = ?

T₁ = 298 K

We input the data:

852 mL * 200 K = V₂ * 298 K

And solve for V₂:

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2 years ago

What volume of 0.152 M KMnO4 solution would completely react with 20.0 mL of 0.381 M FeSO4 solution according to the following n

ANEK [815]

Answer: The volume of permanganate ion (potassium permanganate) is 10.0 mL

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of ferrous sulfate solution = 0.381 M

Volume of solution = 20.0 mL = 0.020 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.381M=\frac{\text{Moles of ferrous sulfate}}{0.020L}\\\\\text{Moles of ferrous sulfate}=(0.381mol/L\times 0.020L)=0.00762mol$

For the given chemical equation:

$5Fe^{2+}+8H^++MnO_4^-\rightarrow 5Fe^{3+}+Mn^{2+}+4H_2O$

By Stoichiometry of the reaction:

5 moles of iron (II) ions (ferrous sulfate) reacts with 1 mole of permanganate ion (potassium permanganate)

So, 0.00762 moles of iron (II) ions (ferrous sulfate) will react with = $\frac{1}{5}\times 0.00762=0.00152mol$ of permanganate ion (potassium permanganate)

Now, calculating the volume of permanganate ion (potassium permanganate) by using equation 1, we get:

Molarity of permanganate ion (potassium permanganate) = 0.152 M

Moles of permanganate ion (potassium permanganate) = 0.00152 mol

Putting values in equation 1, we get:

$0.152mol/L=\frac{0.00152mol}{\text{Volume of permanganate ion (potassium permanganate)}}\\\\\text{Volume of permanganate ion (potassium permanganate)}=\frac{0.00152mol}{0.152mol/L}=0.01L=10.0mL$

Hence, the volume of permanganate ion (potassium permanganate) is 10.0 mL

3 0

3 years ago

What is the conservation of mass

adelina 88 [10]

Answer: This is the definition if that's what you want.

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3 years ago

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satela [25.4K]

Answer:

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Answer:

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Explanation:

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3 years ago