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3 years ago

An excess of hydrogen ions in the body fluids can have fatal results because this can

Chemistry

1 answer:

valentinak56 [21]3 years ago

6 0

Answer:

An excess of hydrogen ions in the body fluids can have disastrous results because. excess hydrogen ions can break chemical bonds. excess hydrogen ions can kill living cells.

Explanation:

It is that because, excess hydrogen ions can change the shape of large complex molecules, rendering them non-functional.

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Calculate the pH at of a 0.10 Msolution of anilinium chloride . Note that aniline is a weak base with a of . Round your answer t

notsponge [240]

The question is incomplete, here is the complete question:

Calculate the pH at of a 0.10 M solution of anilinium chloride $(C_6H_5NH_3Cl)$ . Note that aniline $(C6H5NH2)$ is a weak base with a $pK_b$ of 4.87. Round your answer to 1 decimal place.

Answer: The pH of the solution is 5.1

Explanation:

Anilinium chloride is the salt formed by the combination of a weak base (aniline) and a strong acid (HCl).

To calculate the pH of the solution, we use the equation:

$pH=7-\frac{1}{2}[pK_b+\log C]$

where,

$pK_b$ = negative logarithm of weak base which is aniline = 4.87

C = concentration of the salt = 0.10 M

Putting values in above equation, we get:

$pH=7-\frac{1}{2}[4.87+\log (0.10)]\\\\pH=5.06=5.1$

Hence, the pH of the solution is 5.1

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3 years ago

Complete el crucigrama sobre conceptos del agua.

strojnjashka [21]

Aumento disminuyen el Aguascalientes cambia se estado li

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3 years ago

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riadik2000 [5.3K]

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3 years ago

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N76 [4]

Answer:

Explanation:

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3 years ago

Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr (1)

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = 181,66 kJ/mol

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = 0,216 kJ/Kmol

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ 4. +117,1 kJ/mol

I hope it helps!

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3 years ago