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Fiesta28 [93]
1 year ago
6

a 4.4 g sample of gas occupies 2.24 l of volume at stp. without thinking too hard, what is the mw of the gas, and name two gases

that would have this mw. write out your logic.
Chemistry
1 answer:
garri49 [273]1 year ago
7 0

As a result, gas's molecular weight is 44g/mol. It is the gas's (Carbon dioxide) molecular weight.

What is Molecular Weight?

The total atomic weights of the atoms in a molecule are measured by its molecular weight. To calculate stoichiometry in chemical equations and reactions, chemists employ molecular weight. M.W. or MW are two frequent abbreviations for molecular weight. Atomic mass units (amu), Daltons, or a unitless expression can be used to indicate molecular weight (Da).

The mass of the isotope carbon-12, which is given a value of 12 amu, serves as the reference point for defining both atomic weight and molecular weight. Because there are many carbon isotopes, the atomic weight of carbon is not exactly 12.

A mole of any gas at STP takes up a volume of 22.4l

at STP, a gas fills a volume of 2.24l

Calculating the quantity of moles of gas in step two

Consequently, the amount of gas in moles

= 0.1 moles

This is equivalent to carbon dioxide's molecular weight.

=44g/mol

As a result, gas's molecular weight is 44g/mol. It is the gas's (Carbon dioxide) molecular weight.

Learn more about Molecular Weight from given link

brainly.com/question/837939

#SPJ4

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What is the density of 0.50 grams of gaseous carbon stored under 1.5 atm of pressure at a temperature of -20.0 C?
Colt1911 [192]

Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.

Explanation:

  • d = m/V, where d is the density, m is the mass and V is the volume.
  • We have the mass m = 0.50 g, so we must get the volume V.
  • To get the volume of a gas, we apply the general gas law PV = nRT

P is the pressure in atm (P = 1.5 atm)

V is the volume in L (V = ??? L)

n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).

  • Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
  • Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.
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