Answer:
V= 0.031L
Explanation:
P= 0.97atm, V= ?, n= 0.12/98 =0.00122mol, R= 0.082, T= 22.4+273= 295.4
Applying
PV=nRT
0.970×V = 0.00122×0.082×295.4
Simplify the above equation
V= 0.031L
Answer:
<em><u>The products of the combustion of hydrocarbons are carbon dioxide and water. Many hydrocarbons are used as fuel because their combustion releases very large amounts of heat energy.</u></em>
Answer:
0.049 mol/L.s
Explanation:
The decomposition of hydrogen peroxide is:

![Rate = -\dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{ 2 \Delta [H_2O_2]}{\Delta t}](https://tex.z-dn.net/?f=Rate%20%3D%20-%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%202%20%20%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D)
The rate of decomposition reaction = the rate of formation of
= 0.098 mol/L.s
∴
Rate of formation of


= 0.049 mol/L.s
Answer:
V₂ = 1.41 L
Explanation:
Given data:
Initial temperature = 35°C (35 +273.15 K = 308.15 K)
Initial volume = 1.5 L
Final temperature = 17°C (17+273.15 K = 290.15 K)
Final volume = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
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 = 1.5 L × 290.15 K / 308.15 k
V₂ = 435.23 L.K / 308.15 k
V₂ = 1.41 L