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Anastaziya [24]
11 months ago
6

A 75.0-liter canister contains 15.82 moles of argon at a pressure of 546.8 kilopascals. what is the temperature of the canister?

the temperature of the canister is k.
Chemistry
1 answer:
Eduardwww [97]11 months ago
7 0

The temperature of the canister is 311.82(OR 312)K

Given:

volume of canister = 75.0-liter

moles of argon = 15.82 moles

pressure = 546.8 kilopascals

To Find:

temperature of the canister

Solution: Pressure of argon = 546.8 kPa

Conversion factor: 1 atm = 101.325 kPa

According to Ideal gas law,

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature

T = PV/nR = (5.4 atm x 75.0 L) / (15.82 x 0.0821 L.atm.mol-¹K-¹)

T = 311.82 K

Hence the temperature of the canister is 311.82 K.

Learn more about Temperature here:

brainly.com/question/25677592

#SPJ4

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Density = mass/volume
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How does your Skeletal and Muscular Systems work together?
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How many molecules are in 41.8 g of sulfuric acid
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Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.

Therefore, 0.426187053 moles of sulfuric acid is equal

\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

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A student made this model of a carbon dioxide molecule. its has one carbon atom and two oxygen atoms what is the correct way to
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A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound
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Answer:

The empirical formula of the compound is C_{0.504}HO_{1.008}.

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen (H) is scaled up to a mole, since it has the molar mass, and both carbon (C) and oxygen (O) are also scaled up in the same magnitude. The empirical formula is of the form:

C_{x}HO_{y}

Where x, y are the number of moles of the carbon and oxygen, respectively.

The scale factor (r), no unit, is calculated by the following formula:

r = \frac{M_{H}}{m_{H}} (1)

Where:

m_{H} - Mass of hydrogen, in grams.

M_{H} - Molar mass of hydrogen, in grams per mole.

If we know that  M_{H} = 1.008\,\frac{g}{mol} and m_{H} = 0.2\,g, then the scale factor is:

r = \frac{1.008}{0.2}

r = 5.04

The molar masses of carbon (M_{C}) and oxygen (M_{O}) are 12.011\,\frac{g}{mol} and 15.999\,\frac{g}{mol}, then, the respective numbers of moles are: (r = 5.04, m_{C} = 1.2\,g, m_{O} = 3.2\,g)

Carbon

n_{C} = \frac{r\cdot m_{C}}{M_{C}} (2)

n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }

n_{C} = 0.504\,moles

Oxygen

n_{O} = \frac{r\cdot m_{O}}{M_{O}} (3)

n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }

n_{O} = 1.008\,moles

Hence, the empirical formula of the compound is C_{0.504}HO_{1.008}.

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