Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
We know that,
The relation between the 
for an ideal gas are :
As we are given :
Now we have to calculate the entropy change of the gas.
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Answer:
Iodine has more electron levels (rings) than fluorine.
Explanation:
On the periodic table, fluorine is on period 2 so it has 2 outer rings, iodine is on period 5 so it has 5 outer rings. The more rings the higher radius.
Both acids and bases cause indicator dyes to change colors.
Answer:
Explanation:
<em>Percent ionization</em> is the percent of the original acid that has ionized:
- %, ionization = (molar concentration of hydrogen ions at equilibrium / molar concentration of original acid) × 100
<u><em>Part A:</em></u>
<u>1) Data:</u>
- Ka: 6.7 × 10 ⁻⁷
- [HA] = 0.10 M
- %, ionization = ?
<u>2) Equilibrium equation:</u>
<u>3) ICE (initial, change, equilbirium) table </u>
Concentrations
HA H⁺ A⁻
Initial 0.10 0 0
Change - x + x + x
Equilibrium 0.10 - x x x
- Equation: Ka = [H⁺] [A⁻] / [HA] =
6.7 × 10 ⁻⁷ = x² / (0.10 - x)
<u>4) Solve the equation:</u>
Since Ka << 1, you can assume x << 0.10 and 0.10 - x ≈ 0.10
- 6.7 × 10 ⁻⁷ ≈ x² / 0.10 ⇒ x² ≈ 6.7 × 10⁻⁸ ⇒ x ≈ 2.588 × 10⁻⁴
- % ionization ≈ (2.588 × 10⁻⁴ M / 0.1 M) × 100 ≈ 0.2588 % ≈ 0.26% (two significant figures)
<u><em>Part B:</em></u>
<u>1) Data:</u>
- Ka: 6.7 × 10 ⁻⁷
- [HA] = 0.010 M
- %, ionization = ?
<u>2) Equilibrium equation:</u>
<u>3) ICE table:</u>
Concentrations
HA H⁺ A⁻
Initial 0.010 0 0
Change - x + x + x
Equilibrium 0.010 - x x x
- Equation: Ka = [H⁺] [A⁻] / [HA] =
6.7 × 10 ⁻⁷ = x² / (0.010 - x)
<u>4) Solve the equation</u>:
Since Ka << 1, you can assume x << 0.010 and 0.010 - x ≈ 0.010
- 6.7 × 10 ⁻⁷ ≈ x² / 0.010 ⇒ x² ≈ 6.7 × 10⁻⁹ ⇒ x ≈ 8.185 × 10⁻5
- % ionization ≈ (8.185 × 10⁻⁵ M / 0.010 M) × 100 ≈ 0.8185 % ≈ 0.82% (two significant figures)
