A sample of gas is compressed from 3.25 L to 1.20 L at constant temperature. If the pressure of this gas in the 3.25-L volume is 100.00 kPa, what will the pressure be at 1.20 L?
Answers
Answer:
. The reaction of 50 mL of N2 gas with 150 mL H2 gas to form ammonia via the equation: N 2(g) + 3H2 (g) → 2NH3 (g) will produce a total of mL of ammonia if pressure and temperature are kept constant
Explanation:
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Answer:
The stress of the gasoline at 1.20 L extent would be 270.83 kPa, assuming that the temperature stays constant.
Explanation:
According to Boyle's Law, the strain and extent of a fuel are inversely proportional, as lengthy as the temperature stays constant. This capacity that:
P1V1 = P2V2
where P1 and V1 are the preliminary strain and volume, and P2 and V2 are the remaining stress and volume.
In this case, we know:
P1 = one hundred kPa
V1 = 3.25 L
V2 = 1.20 L
We can use these values to remedy for P2:
P1V1 = P2V2
100.00 kPa × 3.25 L = P2 × 1.20 L
P2 = (100.00 kPa × 3.25 L) / 1.20 L
P2 = 270.83 kPa
Therefore,
The stress of the gasoline at 1.20 L extent would be 270.83 kPa, assuming that the temperature stays constant.
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