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inysia [295]
3 years ago
10

A 5.00 gram sample of magnesium sulfate hydrate (epsom salt) is heated in the lab to form anhydrous magnesium sulfate. After hea

ting the substance, it is recorded that 2.86 grams of anhydrous magnesium sulfate remain. What is the chemical formula for this sample of magnesium sulfate hydrate
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

The answer to your question is  MgSO₄ 5H₂O

Explanation:

Data

mass of MgSO₄ = 2.86 g

mass of H₂O = 2.14 g (5 - 2.86)

Process

1.- Calculate the molecular mass of the compounds

MgSO₄ = 24 + 32 + (16 x 4) = 120

H₂O = 16 + 2 = 18

2.- Convert the grams obtain to moles

                        120 g of MgSO₄  --------------- 1 mol

                         2.8 g                  ----------------  x

                         x = (2.8 x 1)/120

                        x = 0.024 moles

                         18 g of H₂O --------------------- 1 mol

                         2.14 g           -------------------- x

                         x = (2.14 x 1)/18

                         x = 0.119

3.- Divide by the lowest number of moles

MgSO₄  = 0.024/0.024 = 1

H₂O = 0.119/ 0.024 = 5

4.- Write the molecular formula

                   MgSO₄5H₂O                        

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2 years ago
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Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH
Doss [256]

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

    No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{12 g}{60 g/mol}

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

              Density = \frac{mass}{volume}

                 1 g/ml = \frac{100 g}{volume}

                    volume = 100 ml

Hence, molarity = \frac{\text{no. of moles}}{volume}

                           = \frac{0.2 mol}{0.1 L}

                           = 2 mol/l

As reaction equation for the given reaction is as follows.

     NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O

So,          moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

          x \times 1 M = 10 mL \times 2 M     (as 1 L = 1000 ml)

                        x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.                    

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