Answer:
221 °C
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 4.1 L
Initial temperature (T₁) = 25 °C
= 25 °C + 273
= 298 K
Final volume (V₂) = 6.8 L
Final temperature (T₂) =?
The final temperature of the gas can be obtained as follow:
V₁ / T₁ = V₂ / T₂
4.1 / 298 = 6.8 / T₂
Cross multiply
4.1 × T₂ = 298 × 6.8
4.1 × T₂ = 2026.4
Divide both side by 4.1
T₂ = 2026.4 / 4.1
T₂ ≈ 494 K
Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:
°C = K – 273
K = 494
°C = 494 – 273
°C = 221 °C
Thus the final temperature of the gas is 221 °C
Mark the lime poor conducter of heat and electricity.
Yes why no ,Xenon is already has its secured place in modern periodic table.
- It belongs to group 18,the Helium family and inert gas group having atomic no Z=54
Answer:
The equilibrium concentration of NO is 0.02124 M.
Explanation:
Given that,
Initial concentration of NOBr = 0.878 M

Temperature = 24°C
We know that,
The balance equation is

Initial concentration is,

Concentration is,

Equilibrium concentration

We need to calculate the value of x
Using formula of concentration
![k_{c}=\dfrac{[NO][Br_{2}]}{[NOBr]^2}](https://tex.z-dn.net/?f=k_%7Bc%7D%3D%5Cdfrac%7B%5BNO%5D%5BBr_%7B2%7D%5D%7D%7B%5BNOBr%5D%5E2%7D)
Put the value into the formula
![3.07\times10^{-4}=\dfrac{[2x][x]}{[0.878-2x]^2}](https://tex.z-dn.net/?f=3.07%5Ctimes10%5E%7B-4%7D%3D%5Cdfrac%7B%5B2x%5D%5Bx%5D%7D%7B%5B0.878-2x%5D%5E2%7D)





We need to calculate the equilibrium concentration of NO
Using formula of concentration of NO

Put the value of x


Hence, The equilibrium concentration of NO is 0.02124 M.
Answer:
The rate of the reaction increased by a factor of 1012.32
Explanation:
Applying Arrhenius equation
ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)
where;
k₂/k₁ is the ratio of the rates which is the factor
Ea is the activation energy = 274 kJ/mol.
T₁ is the initial temperature = 231⁰C = 504 k
T₂ is the final temperature = 293⁰C = 566 k
R is gas constant = 8.314 J/Kmol
Substituting this values into the equation above;
ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)
ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)
ln(k₂/k₁) = 6.92
k₂/k₁ = exp(6.92)
k₂/k₁ = 1012.32
The rate of the reaction increased by 1012.32
Is a compound that went through a chemicAl change