Answer:
5.9 x 10^-9 N.
Explanation:
Below is an attachment containing the solution.
First, let's assign variables for the moles of CuSO₄·5H₂O as x and moles of MgSO₄·7H₂O as y. The molar mass of the substances are the following
CuSO₄·5H₂O: <span>249.685 g/mol
</span>CuSO₄: <span>159.609 g/mol
</span>MgSO₄·7H₂O: <span>246.49 g/mol
</span>MgSO₄: <span>120.366 g/mol
</span>H₂O: 18 g/mol
The solution is as follows:
2.988 = 249.685x + 246.49y --> eqn 1
5.02 - 2.988 = 18(5x + 7y) --> eqn 2
Solving both equations simultaneously, the values of x and y are:
x = 0.0134 mol CuSO₄·5H₂O
y = 0.0257 mol MgSO₄·7H₂O
Thus, the percent CuSO₄·5H₂O is equal to
Mass Percentage = [(0.0134 mol CuSO₄·5H₂O)*(249.685 g/mol )]/{[(0.0134 mol CuSO₄·5H₂O)*(249.685 g/mol )] + [(0.0257 mol MgSO₄·7H₂O)*(246.49 g/mol)]} * 100
Mass percentage = 34.56%
X * 0.10 mg/mL = 0.50 mg
x = 0.50 mg / 0.10
x = 5.0 mL
hope this helps!
<span>NO2 weighs 46.005 grams per mol. There are 6.02x10^23 molecules in a mol. In the given sample of 189.5 grams, there are 4.12 mols. This means that there are 2.48x10^24 molecules of NO2 in the given sample.</span>