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Dominik [7]
2 years ago
5

How many grams of S are in 475 g of SO2?

Chemistry
1 answer:
enot [183]2 years ago
6 0

There are 237. 5 g of Sulfur,S in 475 g of SO2?

<h3 />

<h3>Calculation of grams of Sulfur</h3>

From the question, we can say that

  • The molar mass of sulfur = 32 g/mol
  • The molar mass of oxygen = 16 g/mol

Therefore,

The molar mass for SO2 = 32 + (16 × 2) g/mol  = 64 g/mol

Now,

If 1 mole of SO2 contains 1 mole of S

Then 64 g of SO2, will contain 32g of S;

Such that

475 g of SO2 will give { \frac{ 32g of S) ( 475 g of SO2)}{64 of SO2} }  = 237. 5 g of Sulfur.

Learn more about molar mass here :brainly.com/question/18291695

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Answer:

2.73

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How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?
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Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,         molarity = \frac{no. of moles}{Volume (in L) of solution}

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

           molarity = \frac{no. of moles}{Volume of solution in liter}

            0.0800 M = \frac{no. of moles}{0.05 L}

            no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

                No. of moles = \frac{mass in grams}{molar mass}

             mass in grams = no. of moles \times molar mass of CuSO_{4}.5H_{2}O

                                       = 1.6 mol \times 249.68 g/mol

                                       = 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

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