The rate constant : k = 9.2 x 10⁻³ s⁻¹
The half life : t1/2 = 75.3 s
<h3>Further explanation</h3>
Given
Reaction 45% complete in 65 s
Required
The rate constant and the half life
Solution
For first order ln[A]=−kt+ln[A]o
45% complete, 55% remains
A = 0.55
Ao = 1
Input the value :
ln A = -kt + ln Ao
ln 0.55 = -k.65 + ln 1
-0.598=-k.65
k = 9.2 x 10⁻³ s⁻¹
The half life :
t1/2 = (ln 2) / k
t1/2 = 0.693 : 9.2 x 10⁻³
t1/2 = 75.3 s
2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
The answer is (3) HClO. In the Cl2, chlorine has an oxidation number of zero. In HCl, the oxidation number is -1. In HClO2, the oxidation number is +3. In HClO, it is +1. You can calculate this by using O with oxidation number of -2 and H with +1.
To find the ratio of the the combination for the ion, write the charge of the cation as the subscript for the anion, and the charge of the anion as the subscript of the cation. This will make the charges effectively cancel and you will be left with a neutral ionic compound. Remember, that an ionic compound is made up of a metal and a nonmetal.
For Ca2+ and Cl-, you will get the neutral compound to be CaCl₂.
