Answer:
5 significant figures
Explanation:
(2) 0.10 M K2SO4(aq)
(3) 0.10 M K3PO4(aq)
(4) 0.10 M KNO3(aq)
Answer:
T₂ = 379.4 K
Explanation:
Given data:
Initial volume = 1.56 L
Initial temperature = 20°C (20+273 = 293 K)
Final volume = 2.02 L
Final temperature = ?
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
T₂ = V₂T₁ /V₁
T₂ = 2.02 L × 293 K / 1.56 L
T₂ = 591.86 L.K / 1.56 L
T₂ = 379.4 K
Which intervals are affected by the addition of a
catalyst?
(1) 1 and 2
(2) 1 and 3
(3) 2 and 4
(4) 3 and 4
Intervals 1 and 3 are most affected by the addition of catalyst. Option 2 is correct.
Catalysts are those compounds that reduce the activation energy for the reaction to occur, hence increasing the rate of reaction.
Activation energy is the energy that is required to start a reaction. Here, 1 represents the activation energy and 3 represents activation energy without catalyst.
Therefore, intervals 1 and 3 are most affected by the addition of catalyst.
Learn more about Catalyst:
Answer: Option (2) is the correct answer.
Explanation:
The minimum amount of energy required by the reactants to undergo a reaction is known as activation energy.
And when we add a catalyst into a reaction then it leads to a decrease in activation energy. As a result, product formation becomes faster.
Therefore, we can conclude that intervals 1 and 3 are affected by the addition of a catalyst.