Answer:
6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.
Explanation:
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given:
Concentration is decreased to 1.56 % which means that 0.0156 of
is decomposed. So,
= 0.0156
Thus,
kt = 4.1604
The expression for the half life is:-
Half life = 15.0 hours
Where, k is rate constant
So,

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>
Answer: The molar mass of the gas is 31.6 g/mol
Explanation:
According to ideal gas equation:
P = pressure of gas = 3.0 atm
V = Volume of gas = 25.0 L
n = number of moles = ?
R = gas constant =
T =temperature =
Moles =


The molar mass of the gas is 31.6 g/mol
Answer:
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Explanation:
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Explanation: This is a reaction of oxidation of
in the presence of acidified
. Acidified
is a strong oxidizing agent.
To balance out the
on the reactant side, we write
on the product side.
Balancing out the following reaction gives us:

Answer:
Water has the greatest ΔEN
ΔEN H₂O → 3.4 - 2.1 = 1.3 Option D.
Explanation:
We should find the Electronegativity data in the Periodic table for all the elements:
C : 2.6
O: 3.4
H: 2.1
S: 2.6
N: 3.0
a. ΔEN CO₂ → 3.4 - 2.6 = 0.4
b. ΔEN H₂S → 2.6 - 2.1 = 0.5
c. ΔEN NH₃ → 3 - 2.1= 0.9
d. ΔEN H₂O → 3.4 - 2.1 = 1.3