Answer:
35.29%
Explanation:
% error = |experimental value - Accepted value| / Accepted Value x 100
Number of questions: 34
Number of questions answered: 22
%error = |22-34| / 34 x 100
Hence, John's percentage error is 35.29 % or 35.3%
H = 1 amu
P = 31 amu
O = 16 amu
Therefore:
H3PO4 = 1 x 3 + 31 + 16 x 4 => 98 u
hope this helps!
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.
If you want an example then a car passing by a standing man is the best one !!
Answer:
ΔpH = 0.20
Explanation:
The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2
HCO₃⁻ ⇄ H⁺ + CO₃²⁻
There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.
Using Henderson-Hasselbalch formula:
pH = pka + log [Base] / [Acid]
pH = 10.2 + log 0.342mol / 0.479mol
<em>pH = 10.05</em>
NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:
NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺
0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:
CO₃²⁻: 0.342mol + 0.091mol: 0.433mol
HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol
Using Henderson-Hasselbalch formula:
pH = pka + log [Base] / [Acid]
pH = 10.2 + log 0.433mol / 0.388mol
pH = 10.25
That means change of pH, ΔpH is:
ΔpH = 10.25 - 10.05 = <em>0.20</em>
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I hope it helps!
