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fgiga [73]
3 years ago
8

A can of coke contains 25 mL of carbon dioxide gas at 100kPa. If you take it on a hike up Mount Everest and the pressure decreas

es to 50 kPa, what will the new volume of the carbon dioxide gas in your coke can be?
Chemistry
1 answer:
Mazyrski [523]3 years ago
7 0

Answer:

50 mL

Explanation:

We can solve this problem by using Boyle's Law, which states that:

"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

p\propto \frac{1}{V}

where

p is the pressure of the gas

V is the volume of the gas

The equation can be rewritten as

p_1 V_1 = p_2 V_2

where in this problem:

p_1 = 100 kPa is the initial pressure of the gas in the coke

V_1=25 mL is the initial volume

p_2=50 kPa is the final pressure

Solving for V2, we find the final volume:

V_2=\frac{p_1 V_1}{p_2}=\frac{(100)(25)}{50}=50 mL

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28.7 l of propane, c3h8, are consumed in a combustion reaction. how many grams of water are produced?
lora16 [44]
The chemical reaction would be:

C3H8 + 5O2 = 3CO2 + 4H2O

For this case, we assume that gas is ideal thus in every 1 mol the volume would be 22.41 L. We calculate as follows:

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Hope this answers the question.
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All organic molecules contain carbon and hydrogen. what parts of an organic molecule may contain oxygen, nitrogen, or phosphorus
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3 years ago
A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.
kondaur [170]

The question is incomplete, complete question is:

A fluid occupying 3.2 m^3 of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = 3.2 m^3

Density of the fluid = D

D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3

Specific volume is the reciprocal of the density :

V_{specific}=\frac{1}{Density}

Specific volume of the fluid = S_v

S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg

2)

Density of the fluid in English Engineering units  = D (lb/ft^3)

1 kg = 2.20462 lb

1 m = 3.280 ft

D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3

Specific volume of the fluid :

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3)

Density of the fluid in British Gravitational System units  = D (slug/ft^3)

1 kg = 0.06852 slug

1 m = 3.280 ft

D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3

Specific volume of the fluid :

=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug

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