Answer:
pH = 6.999
The solution is acidic.
Explanation:
HBr is a strong acid, a very strong one.
In water, this acid is totally dissociated.
HBr + H₂O → H₃O⁺ + Br⁻
We can think pH, as - log 7.75×10⁻¹² but this is 11.1
acid pH can't never be higher than 7.
We apply the charge balance:
[H⁺] = [Br⁻] + [OH⁻]
All the protons come from the bromide and the OH⁻ that come from water.
We can also think [OH⁻] = Kw / [H⁺] so:
[H⁺] = [Br⁻] + Kw / [H⁺]
Now, our unknown is [H⁺]
[H⁺] = 7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]
[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) / [H⁺]
This is quadratic equation: [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴
a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴
(-b +- √(b² - 4ac) / (2a)
[H⁺] = 1.000038751×10⁻⁷
- log [H⁺] = pH → 6.999
A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.
Answer:
1.47 atm
Explanation:
Step 1: Given data
- Initial volume (V₁): 32.4 L
- Initial pressure (P₁): 1 atm (standard pressure)
- Initial temperature (T₁): 273 K (standard temperature)
- Final volume (V₂): 28.4 L
- Final temperature (T₂): 352 K
Step 2: Calculate the final pressure of the gas
We can calculate the final pressure of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
P₂ = P₁ × V₁ × T₂ / T₁ × V₂
P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm
Answer: 600 mL
Explanation:
Given that;
M₁ = 5.85 m
M₂ = 1.95 m
V₁ = 200 mL
V₂ = ?
Now from the dilution law;
M₁V₁ = M₂V₂
so we substitute
5.85 × 200 = 1.95 × V₂
1170 = 1.95V₂
V₂ = 1170 / 1.95
V₂ = 600 mL
Therefore final volume is 600 mL
<span>he specific heat capacity of liquid water is 4.186 J/gm K.</span>
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