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Ymorist [56]
3 years ago
6

A lab technician needs to create 570.0 milliliters of a 2.00 M solution of magnesium chloride (MgCl2). To make this solution, ho

w many grams of magnesium chloride does the technician need?
Refer to the periodic table for help. Express your answer to three significant figures.
Chemistry
1 answer:
love history [14]3 years ago
7 0

Answer:

108.3g

Explanation:

Given parameters:

Volume of solution = 570mm = 0.57L (1000mm = 1L)

Molarity of solution = 2M

Unknown

Mass of MgCl₂

Solution

We first find the  number of moles in the given concentration of magnessium chloride using the expression below:

       Number of moles of MgCl₂= Molarity x volume = 2 x 0.57 = 1.14moles

Using this known moles, we can the unknown mass of MgCl₂ the technician would require:

       Mass of MgCl₂ required = number of moles of MgCl₂ x molar mass

Molar mass of MgCl₂:

Atomic mass of Cl = 35.5g

Atomic mass of Mg = 24g

MgCl₂ = 24 + (35.5x2) = 95gmol⁻¹

Mass of MgCl₂ required = 1.14mole x 95gmole⁻¹ = 108.3g

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Answer:

The answer to your question is:

Explanation:

Data

carbon        7.3%          =     7.3g

hydrogen    4.5%         =      4.5g

oxygen       36.4%         =     36.4 g

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Now

For carbon

                    12 g --------------------1 mol

                    7.3 g     -------------     x

                       x = 7.3/12 = 0.608 mol

For hydrogen

                 1 g   --------------------  1 mol

                 4.5 g  ------------------    x

                   x = 4.5 mol

For oxygen

             16 g ------------------- 1 mol

             36.4 g ----------------    x

             x = 2.28 mol

For nitrogen

              14 g   ----------------   1 mol

              31.8 g ---------------    x

             x = 2.27 mol

Now divide by the lowest result, the is 0.608 from carbon

carbon              0.608/0.608 = 1

hydrogen           4.5/ 0.608 = 7.4

oxygen              2.28/0.608 = 3.75

nitrogen             2.27/0.608 = 3.73

Empirical formula = CH₇O₄N₄

     

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The ka of hf is 6.8 x 10-4. what is the ph of a 0.35 m solution of hf?
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When the reaction equation is:

HF ↔ H+   +   F-

and when the Ka expression
= concentration of products/concentration of reactions

so, Ka = [H+][F-]/[HF]

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and [HF] = 0.35 - X

So, by substitution:

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