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3 years ago

The mass of 3.45 L of hydrogen gas at SATP

Chemistry

1 answer:

Licemer1 [7]3 years ago

5 0

Answer:

Explanation:

The easy way is knowing that all gases basically are 22.4 L/mol at SATP.

There fore 3.45 L/22.4 L/mol =0.154 mol

Or use PV=nRT

n=(1 atm)(3.45L)/(273 K)(0.82 L atm)/(mol K) =0.154 mol

same answer

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3 years ago

What volume (ml) of fluorine gas is required to react with 1. 28 g of calcium bromide to form calcium fluoride and bromine gas a

Alexandra [31]

144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

<h3>What is Ideal Gas Law ? </h3>

The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.

PV = nRT

where,

P = Presure

V = Volume in liters

n = number of moles of gas

R = Ideal gas constant

T = temperature in Kelvin

Here,

P = 1 atm [At STP]

R = 0.0821 atm.L/mol.K

T = 273 K [At STP]

Now first find the number of moles

F₂ + CaBr₂ → CaF₂ + Br₂

Here 1 mole of F₂ reacts with 1 mole of CaBr₂.

So, 199.89 g CaBr₂ reacts with = 1 mole of F₂

1.28 g of CaBr₂ will react with = n mole of F₂

$n = \frac{1.28\g \times 1\ \text{mole}}{199.89\ g}$

n = 0.0064 mole

Now put the value in above equation we get

PV = nRT

1 atm × V = 0.0064 × 0.0821 atm.L/mol.K × 273 K

V = 0.1434 L

V ≈ 144 mL

Thus from the above conclusion we can say that 144 mL of fluorine gas is required to react with 1.28 g of calcium bromide to form calcium fluoride and bromine gas at STP.

Learn more about the Ideal Gas here: brainly.com/question/20348074

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