Answer:
The pressure decrease with the passage of time due to again dissolution of carbon-dioxide gas in the liquid solution.
Explanation:
In soft drinks, the carbon-dioxide gas is added in the drinks with high pressure because carbon-dioxide is a gas which is insoluble in soft drink at room temperature but soluble in the drinks at high pressure so when the pressure is removed from the soft drink, the carbon-dioxide gas releases in the air with the passage of time. But in close bottle , there is no place of escape so it again dissolve in the solution.
Answer:
V₂ = 2.91 L
Explanation:
Given data:
Initial volume = 3.50 L
Initial temperature = 90.0°C (90+273 = 363 K)
Final temperature = 30.0 °C ( 30 +273 = 303 K)
Final volume = ?
Solution:
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
V₁/T₁ = V₂/T₂
3.50 L / 363 K) = V₂ / 303 K)
V₂ = 0.0096 L/K × 303 K
V₂ = 2.91 L
Answer:
2H2S + 3O2 → 2SO2 + 2H2O
Explanation:
Step 1: Data given
Hydrogen sulfide = H2S
Oxygen = O2
sulfur dioxide = SO2
water = H2O
Step 2: The unbalanced equation
H2S + O2 → SO2 + H2O
Step 3: Balancing the equation
H2S + O2 → SO2 + H2O
On the left side we have 2x O (in O2) and on the right side we have 3x O (2x in SO2 and 1x in H2O). To balance the amount of O, we have to multiply O2 (on the left side) by 3 and SO2 and H2O on the right side by 3.
H2S + 3O2 → 2SO2 + 2H2O
On the right side we have 4x H and on the left side we have 2x H. To balance the amount of H, we have to multiply H2S by 2.
Now the equation is balanced.
2H2S + 3O2 → 2SO2 + 2H2O
