Answer:
pH = 10
The solution is basic.
Explanation:
A solution contains 1 × 10⁻⁴ M OH⁻ ions. First, we will calculate the pOH.
pOH = -log [OH⁻]
pOH = -log 1 × 10⁻⁴
pOH = 4
We can find the pH of the solution using the following expression.
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 4 = 10
Since the pH > 7, the solution is basic.
For the 1st order reactions,rate constant (k) is mathematically expressed as
k =

where, t = time
Co = initial conc. of reactant
Ct = conc. of reactant after time 't'
Given: k = <span>2.20 × 10^-5 s-1, t = 2 hours = 7200 s
Therefore, we have
</span>2.20 × 10^-5 =

∴

= 0.06877
∴,

= 1.1716
∴, Ct = 85.35%
Thus, <span>
85.35 % of the initial amount of SO2Cl2 will remain after 2.00 hours.</span>
Answer:
<u>132.15</u>
Explanation:
Molar mass N = 14.00
Molar mass H = 1.01
Molar mass H4 = 1.01 x 4 = 4.04
Molar mass NH4 = 14.00 + 4.04 = 18.04
Molar mass (NH4)2 = 18.04 x 2 = 36.08
Molar mass S = 32.07
Molar mass O = 16.00
Molar mass O4 = 16.00 x 4 = 64.00
Molar mass SO4 = 32.07 + 64.00 = 96.07
Molar mass (NH4)2SO4 = 36.08 + 96.07 = <u>132.14</u>
Answer:
The equilbrium constant is 179.6
Explanation:
To solve this question we can use the equation:
ΔG = -RTlnK
<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>
<em>R is gas constant = 8.314x10⁻³kJ/molK</em>
<em>T is absolute temperature = 298K</em>
<em>And K is equilibrium constant.</em>
Replacing:
12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK
5.19 = lnK
e^5.19 = K
179.6 = K
<h3>The equilbrium constant is 179.6</h3>