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2 years ago

How many milliliters of this solution are required to prepare 1.00 L of 0.130 M K2Cr2O7 ?

Chemistry

1 answer:

ra1l [238]2 years ago

7 0

Answer:

First you'll need to divide the mass of K2Cr2O7 by its molar mass and volume which should be easy but I'll do some of it to help you out

Explanation:

$K_{2}Cr_{2}O_{7}$

$2K = 39.098 = 2(39.098) = 78.196$

$2Cr = 51.996 = 2(51.996) = 103.992$

and

$7O = 15.999 = 7(15.999) = 111.993$

$78.196 + 103.992 + 111.993 = 294.181$

so that's the mass of K2Cr2O7 and then you divide it by the molar mass and volume.

Hopefully you can do that on your own! Hope I helped a bit atleast and I have a good feeling about 0.438 but I don't want you to just guess

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If 30.0 mL of Ca(OH)2 with an unknown concentration is neutralized by 40.0 mL of 0.175 M HCl, what is the concentration of the C

Over [174]

Answer:

Concentration of Ca(OH)₂:

0.117 M.

Explanation:

How many moles of HCl is consumed?

Note the unit of concentration: moles per liter solution.

$c(\text{HCl}) = 0.175\;\text{M} = 0.175\;\text{mol}\cdot\textbf{L}^{-1}$ .

Convert milliliters to liters.

$V(\text{HCl})=40.0\;\text{mL} = 0.0400\;\text{L}$ .

$n(\text{HCl}) = c(\text{HCl})\cdot V(\text{HCl})= 0.175\;\text{mol}\cdot\text{L}^{-1} \times 0.0400\;\text{L}= 7.00\times 10^{-3}\;\text{mol}$ .

How many moles of NaOH in the solution?

Refer to the equation. The coefficient in front of Ca(OH)₂ is 1. The coefficient in front of HCl is 2. In other words, it takes two moles of HCl to neutralize one mole of Ca(OH)₂. That $7.00\times 10^{-3}\;\text{mol}$ of HCl will neutralize only half that much Ca(OH)₂.

$\displaystyle n(\text{Ca}(\text{OH})_2)=\frac{1}{2}\;n(\text{HCl}) = 3.50\times 10^{-3}\;\text{mol}$ .

What's the concentration of the Ca(OH)₂ solution?

Concentration is the number of moles of solute per unit volume.

$\displaystyle c(\text{Ca}(\text{OH})_2) = \frac{n(\text{Ca}(\text{OH})_2)}{V(\text{Ca}(\text{OH})_2)} = \frac{3.50\times 10^{-3}\;\text{mol}}{0.0300\;\text{L}}=0.117\;\text{mol}\cdot\text{L}^{-1}$ .

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