Which step will decrease the pressure of a gas inside a closed cubical container? increasing the number of moles of gas decreasi
2 answers:
Answer: Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.
Explanation:
According to Gay-Lussac's Law : 'The pressure of the gas increases with increase in temperature of the gas when volume of the gas is kept constant'.
At constant volume, pressure of the gas will decrease on decreasing the temperature or vice versa.
Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.
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Answer:
a) [A⁻]/[HA] = 0.227
b) [A⁻]/[HA] = 0.991
c) [A⁻]/[HA] = 2.667
Explanation:
In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:
- CH₃CH₂CO₂H = HA
- CH₃CH₂CO₂⁻ = A⁻
pH = pka + Log [A⁻]/[HA]
pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]
- (a)
4.23 = 4.874 + Log [A⁻]/[HA]
-0.644 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.227 = [A⁻]/[HA]
- (b)
4.87 = 4.874 + Log [A⁻]/[HA]
-0.004 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.991 = [A⁻]/[HA]
- (c)
5.30 = 4.874 + Log [A⁻]/[HA]
0.426 = Log [A⁻]/[HA]
= [A⁻]/[HA]
2.667 = [A⁻]/[HA]