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3 years ago

8

A student is filtering a mixture of sand and salt water into a beaker. What will be found in the beaker after the filtration is

completed?

2 answers:

Sergio [31]3 years ago

3 0

The student would find the water and sand, because salt dissolves in water unless it was ocean water or sea water

zhuklara [117]3 years ago

3 0

Only salt and sand.
This is because the water would evaporate during the filtration process, leaving only sand and the salt it left behind.
Hope this helps :)

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What is the pH of 0.000134 M solution of HCI?

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Answer: The pH will be 3.87

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.000134$ moles of $HCl$ gives = $\frac{1}{1}\times 0.000134=0.0001342$ moles of $H^+$

Putting in the values:

$pH=-\log[0.000134]$

$pH=3.87$

Thus the pH will be 3.87

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Answer:

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2 years ago

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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

<u>Answer:</u> The value of $K_c$ for the given reaction is 1.435

<u>Explanation:</u>

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

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The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles

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<u>Answer:</u> The expression for equilibrium constant is $K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}$

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

For the general chemical equation:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_c$ is given as:

$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical reaction:

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The expression for $K_{eq}$ is given as:

$K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}$

The concentration of solid is taken to be 0.

So, the expression for $K_{eq}$ is given as:

$K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}$

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Answer:

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