Respuesta :
Explanation:
a) Manganese (II) phosphite :[tex]Mn_3(PO_3)_2[/tex]
Number of phosphite ions in manganese (II) phosphite = [tex]7.23\times 10^{22}[/tex]
In one molecule of manganese (II) phosphite there are 2 ions of phosphite.
Then [tex]7.23\times 10^{22}[/tex] ions of phosphite will be ions will be in:
[tex]\frac{1}{2}\times 7.23\times 10^{22}=1.446\times 10^{23} molecules[/tex]
Moles of manganese (II) phosphite;= [tex]\frac{1.446\times 10^{23}}{6.022\times 10^{23}}= 0.2401 mol[/tex]
Mass of 0.2401 mol of Manganese (II) phosphite :
0.2401 mol × 322.75 g/mol = 77.49 g
77.49 is the mass in grams of a sample of manganese (II) phosphite.
b) Molar mass of ammonium chromate =252.07 g/mol
Percentage of Nitrogen:
[tex]\frac{2\times 14g/mol}{252.07 g/mol}\times 100=11.10\%[/tex]
Percentage of hydrogen:
[tex]\frac{8\times 1g/mol}{252.07 g/mol}\times 100=3.17\%[/tex]
Percentage of chromium:
[tex]\frac{2\times 52 g/mol}{252.07 g/mol}\times 100=41.25\%[/tex]
Percentage of oxygen:
[tex]\frac{7\times 16g/mol}{252.07 g/mol}\times 100=44.44\%[/tex]
c) Molar mass of the substance = 202.23 g/mol
Percentage of the hydrogen = 4.98 %
Let the molecular formula be [tex]C_xH_y[/tex]
Percentage of the carbon = 95.02 %
[tex]\frac{12 g/mol\times x}{202.23 g/mol}\times 100=95.02\%[/tex]
x= 16.01 ≈ 16
Percentage of the hydrogen= 4.98 %
[tex]\frac{1 g/mol\times y}{202.23 g/mol}\times 100=4.98\%[/tex]
y= 10.07 ≈ 10
The molecular formula of the substance is [tex]C_{16}H_{10}[/tex].
The empirical formula of the substance is [tex]C_{\frac{16}{2}}H_{\frac{10}{2}}=C_8H_5[/tex].
