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Oduvanchick [21]
3 years ago
15

HELP!!!!!What is the molar mass of a gas which has a density of .249 g/L at 20.0 degrees Celsius and a pressure of .95 atm?

Chemistry
1 answer:
fgiga [73]3 years ago
7 0

Answer:

6.31g/mol

Explanation:

Using the ideal gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Mole (n) = mass (m)/molar mass (Mm)

* Mm = m/n

Also, density (p) = mass (m) ÷ volume (V)

PV = nRT

Since n = M/Mm

PV = M/Mm. RT

PV × Mm = m × RT

Divide both sides by V

P × Mm = m/V × RT

Since p = m/V

P × Mm = p × RT

Mm = p × RT/P

Mm = 0.249 × 0.0821 × 293/0.95

Mm = 5.989 ÷ 0.95

Mm = 6.31g/mol

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if a volume of air occupying 12.0Lat 20celsius is heated to a new temperature of 100 celsius,what would be the new volume
emmasim [6.3K]

Answer:

V₂ = 15.3

Explanation:

Given data:

Initial volume = 12.0 L

Initial temperature = 20°C

Final temperature =100°C

Final volume = ?

Solution:

First of all we will convert the temperature into kelvin.

20°C + 273 = 293 K

100°C + 273 = 373 K

Formula:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 12.0 L × 373 K / 293 k

V₂ = 4476 L.K /293 k

V₂ = 15.3

V₂ = 1566 L.K / 298 K

V₂ = 5.3 L

6 0
3 years ago
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Answer:

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Explanation:

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In what ways are the two designs in part C different? How do you think these differences will relate to the ability of each desi
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If a pork roast must absorb
Black_prince [1.1K]
Let us  assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
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