Answer:
HI and I-
Explanation:
Hello there!
In this case, according to the acid ionization of hydroiodic acid:
Now, since the acid, HI, results in the conjugate base, I- and the base, H2O in the conjugate acid, H3O+, it is correct to assert that the acid-conjugate base pair is HI and I-
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Answer:
3
Explanation:
The first election shell can only hold 2 electrons, but the next one can hold up to 8
A student measures the volume of a solution to be 0.01370 have 5 significant digits in this measurement.
<h3>What are significant digits?</h3>
The significant digits are the minimum number from zero to nine for reporting any measurement where the digits are uncertain.
The significant digits starting from zero are not significant digits, decimal is not a significant digit, and ending zero after the decimal are significant digits.
Therefore, the student measures the volume of a solution to be 0.01370 5 significant digits are in this measurement.
Learn more about significant digits, here:
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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.
