Answer:
Explanation:
M(s) → M (g ) + 20.1 kJ --- ( 1 )
X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )
M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )
( 3 ) - 2 x ( 2 ) - ( 1 )
M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ
0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ
4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ
heat of formation of M X₄ (g ) is - 773.4 kJ
Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole
Answer is: 79.8 grams of copper(II) sulfate.
N(CuSO₄) = 3.01·10²³; number of molecules.
n(CuSO₄) = N(CuSO₄) ÷ Na.
n(CuSO₄) = 3.01·10²³ ÷ 6.02·10²³ 1/mol.
n(CuSO₄) = 0.5 mol; amount of substance.
m(CuSO₄) = n(CuSO₄) · M(CuSO₄).
m(CuSO₄) = 0.5 mol · 159.6 g/mol.
m(CuSO₄) = 79.8 g; mass of substance.
M - molar mass.
When drawing lewis dot diagram structure we consider the number of valence electrons of the atom.Se has six valence electrons since it is in group six thus it Lewis dot diagram is as follows
..
: Se:
Hydrogen is in group one hence has one valence electron.The lewis dot diagram for 2H is therefore
H:H
The dependent variable is what is being tested. independent is what forces or effects the dependent
Answer:
Mass = 36.38 g
Explanation:
Given data:
Mass of hydrogen = 30.0 g
Mass of nitrogen = 30.0 g
Theoretical yield of reaction = ?
Mass of ammonia = ?
Chemical equation:
N₂+ 3H₂ → 2NH₃
Number of moles of N₂:
Number of moles = Mass /molar mass
Number of moles = 30 g / 28 g/mol
Number of moles = 1.07 mol
Number of moles of H₂:
Number of moles = Mass /molar mass
Number of moles = 30 g / 2g/mol
Number of moles = 15 mol
Now we will compare the moles of NH₃ with hydrogen and nitrogen .
H₂ : NH₃
3 : 2
15 : 2/3×15= 10 mol
N₂ : NH₃
1 : 2
1.07 : 2 × 1.07= 2.14 mol
The number of moles of NH₃ produced by nitrogen are less it will be limiting reactant.
Mass of ammonia:
Mass of NH₃ = moles × molar mass
Mass =2.14 mol × 17 g/mol
Mass = 36.38 g
