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Oduvanchick [21]
2 years ago
12

How many grams of solid Ca(OH)2 (74.1 g/mol) are required to make 500 ml of a 3 M solution?

Chemistry
1 answer:
mars1129 [50]2 years ago
3 0

Answer:

111.15 g are required to prepare 500 ml of a 3 M solution

Explanation:

In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.

Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of

(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.

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5 0
3 years ago
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What is the concentration of a 1:10 dilution of a 2.5 M solution of NaCl? How would you prepare exactly 100 ml of such a solutio
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<u>Answer:</u> The concentration of the solution is 0.25 M

<u>Explanation:</u>

Let the volume of solution of 2.5 M NaCl be 10 mL

We are given:

Dilution ratio = 1 : 10

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

M_1V_1=M_2V_2

where,

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We are given:

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6 0
3 years ago
Consider the reaction 2 al + Fe2O3 to 2Fe + Al2O3. If 60.0g of Al is reacted with excess Fe2O3, determine the amount (in moles)
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    <em><u>calculation</u></em>

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    Step 2: use the mole ratio to determine the  moles of Al2O3.

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Step 3:    finds the mass  of  Al2O3  by us of  <u><em>mass= moles x molar mass</em></u><em> </em>formula.

The molar  mass of Al2O3  =  (2x27)  +( 16 x3) = 102  g/mol

mass is therefore=  102  g/mol  x 1.11= 113.22 grams


             

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