Answer:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2
Ionic bonds are formed when a cation and an anion transfer electrons. The anion gains electrons from the cation to finish its shell, and is usually a nonmetal or a metalloid. A cation gives the anion its electrons to get rid of its partial shell. Cations are metals.
Answer:
V = 12.5 L
Explanation:
Given data:
Volume of NO = 15.0 L
Temperature and pressure = standard
Volume of nitrogen gas produced = ?
Solution:
Chemical equation:
6NO + 4NH₃ → 5N₂ + 6 H₂O
Number of moles of NO:
PV = nRT
n = PV/RT
n = 1 atm × 15.0 L / 0.0821 atm.L /mol.K × 273.15 K
n = 15.0 atm.L / 22.43 atm.L /mol
n = 0.67 mol
now we will compare the moles of No and nitrogen gas.
NO : N₂
6 : 5
0.67 : 5/6×0.67 = 0.56
Volume of nitrogen gas:
PV = nRT
1 atm × V = 0.56 mol × 0.0821 atm.L /mol.K × 273.15 K
V = 12.5 atm.L / 1 atm
V = 12.5 L
The answer is C Does it propaget
J=75. Because he has already used 45 of his/her 120
