Answer:
k ≈ 9,56x10³ s⁻¹
Explanation:
It is possible to solve this question using Arrhenius formula:

Where:
k1: 1,35x10² s⁻¹
T1: 25,0°C + 273,15 = 298,15K
Ea = 55,5 kJ/mol
R = 8,314472x10⁻³ kJ/molK
k2 : ???
T2: 95,0°C+ 273,15K = 368,15K
Solving:



<em>k ≈ 9,56x10³ s⁻¹</em>
I hope it helps!
Answer:
2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid
Explanation:
Write down the equation in the beginning with reactants and products:
NaOH + H₂SO₄ → Na₂SO₄ + H₂0
Now try to balance it. Try with Na first:
2NaOH + H₂SO₄ → Na₂SO₄ + H₂0
Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0
Check if H atoms are also balanced. They are. That means our final reaction is:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0
2 Moles of NaOH reacts with 1 mole of H₂SO₄
Answer:
the Molar heat of Combustion of diphenylacetylene
= 
Explanation:
Given that:
mass of diphenylacetylene
= 0.5297 g
Molar Mass of diphenylacetylene
= 178.21 g/mol
Then number of moles of diphenylacetylene
= 
= 
= 0.002972 mol
By applying the law of calorimeter;
Heat liberated by 0.002972 mole of diphenylacetylene
= Heat absorbed by
+ Heat absorbed by the calorimeter
Heat liberated by 0.002972 mole of diphenylacetylene
= msΔT + cΔT
= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C
= 17756.48 J + 2842.39 J
= 20598.87 J
Heat liberated by 0.002972 mole of diphenylacetylene
= 20598.87 J
Heat liberated by 1 mole of diphenylacetylene
will be = 
= 6930979.139 J/mol
= 6930.98 kJ/mol
Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene
= 
The question is incomplete, here is the complete question:
At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

The reaction is second order for
with a rate constant of
at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M
a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34
<u>Answer:</u> The time taken is 5.19 seconds
<u>Explanation:</u>
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = ?
[A] = concentration of substance after time 't' = 0.150 M
= Initial concentration = 0.260 M
Putting values in above equation, we get:

Hence, the time taken is 5.19 seconds