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3 years ago

MgCl2 + NaOH → Mg(OH)2 +Naci Balance

Chemistry

1 answer:

Oksana_A [137]3 years ago

3 0

MgCl2+2NaOH=Mg(OH)2+2NaCl

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What characteristics are used to clarify an area as a wetland?

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Soil covered, saturated, submerged, flooded w water, standing water

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A plate of the steel hull of a ship develops a large crack when the ship strikes a reef. Analysis of the steel’s composition sho

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Read 2 more answers

In order to gain one pound of body weight, the average person must consume 3500 more calories

Fed [463]

Answer:

17500 calories of chocolate bars are needed to eat to gain 5 pounds.

Explanation:

We can use ratios to calculate the answer using the information given in the question.

1 pound : 3500 grams

5 pounds : x grams

As it is given that the individual is burning no calories, we do not have to factor in any additional numbers.

Method A:

To go from 1 in the first ratio to 5 in the second ratio, they multipled 1 by 5. Hence, to go from 3500 in the first ratio to x in the second ratio, we must multiply by 5.

x = 3500 × 5

x = 17500

Method B:

To solve for the answer x, we can convert the ratios into fractions.

1 / 5 = 3500 / x

3500 / x = 1 / 5

To make x the subject, multiply the denominator of the left fraction with the numerator of the right fraction and place it on the left side. Then multiply the numerator of the left fraction with the denominator of the right fraction and place it on the right side.

x = 5 × 3500

x = 17500

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3 years ago

Limitation for osmosis experiment

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2 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago