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2 years ago

What is the percent composition of the following KBr LiOH Mgl2 Al2(SO4)3

Chemistry

1 answer:

schepotkina [342]2 years ago

4 0

For the first one, K is 32.85% and Br is 67.14%

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3 years ago

Acts as the digestive system inside a cell. It helps to break down old or unneeded parts of the cell, and substances that have b

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2 years ago

If you dilute 19.0 mL of the stock solution to a final volume of 0.310 L , what will be the concentration of the diluted solutio

Dahasolnce [82]

Answer:

$M_2=0.613M_1$

Explanation:

$M_1$ = Concentration of stock solution

$M_2$ = Concentration of solution

$V_1$ = Volume of stock solution = 19 mL

$V_2$ = Volume of solution = 0.31 L= 310 mL

We have the relation

$M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{M_119}{310}\\\Rightarrow M_2=M_1\times\dfrac{19}{310}\\\Rightarrow M_2=0.613M_1$

$\boldsymbol{\therefore M_2=0.613M_1}$

The concentration of the diluted solution will be 0.613 times the concentration of the stock solution.

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2 years ago

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1 year ago

The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was

Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

$[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}$

$[Ca^{2+}]=0.007118 \ M$

$[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}$

$=0.001994 \ M$

Mass of Ca²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}$

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}$

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

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2 years ago

What is the percent composition of the following *KBr *LiOH *Mgl2 *Al2(SO4)3

What is the percent composition of the following KBr LiOH Mgl2 Al2(SO4)3