The answer is -3, if you are asking for that
The atoms of hydrogen that are present in 7.63 g of ammonia(NH3)
find the moles of NH3 =mass/molar mass
7.63 g/ 17 g/mol = 0.449 moles
since there is 3 atoms of H in NH3 the moles of H = 0.449 x 3 = 1.347 moles
by use of 1 mole = 6.02 x10^23 atoms
what about 1.347 moles
= 1.347 moles/1 moles x 6.02 x10^23 atoms = 8.11 x10^23 atoms of Hydrogen
Answer:
Last Quarter also called Third Quarter.
Explanation:
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%