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3 years ago

A solution has a [H+] of 0.20 M. Find the pH of this solution.

Chemistry

1 answer:

Rus_ich [418]3 years ago

4 0

The pH of this solution : 0.699

<h3>Further explanation </h3>

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.

pH = - log [H⁺]

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.

A solution whose value is different by n has a difference in the concentration of H⁺ ion of 10ⁿ.

A solution has a [H+] of 0.20 M, so [H⁺]=0.2

$\tt pH=-log[H^+]\\\\pH=-log[0.2]\\\\pH=1-log~2\\\\pH=0.699$

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3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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4 years ago

A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilib

DochEvi [55]

Answer:

[IBr] = 0.049 M.

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

$I_2+Br_2\rightarrow 2IBr$

It is possible to set up the following equilibrium expression:

$K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110$

Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of $x$ (reaction extent) would be:

$0.0110=\frac{(2x)^2}{(0.50-x)^2}$

Which can be solved for $x$ to obtain two possible results:

$x_1=-0.0277M\\\\x_2=0.0245M$

Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:

$[IBr]=2x=2*0.0249M=0.049M$

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2 years ago

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lakkis [162]

Answer:

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3 years ago

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Answer:

D. The presence of a base catalyst.

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Explanation:

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3 years ago

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GuDViN [60]

It would be NaOH + HCl → <span>NaCl + H2O
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3 years ago