Hydrogen is a non-polar gas with very weak intermolecular forces of attraction. Hydrogen will deviate from the ideal gas behavior at high pressure.
Answer:
CH2N2
Explanation:
To find the molecular formula, we must first find the empirical formula as follows:
28.57% C - 28.57g of Carbon
4.80% H - 4.80g of Hydrogen
66.64% N - 66.64g of Nitrogen
Next, we convert this mass values to mole by dividing by their respective atomic mass.
C = 28.57/12 = 2.38mol
H = 4.80/1 = 4.80mol
N = 66.64/14 = 4.76mol
Next, we divide each mole value by the smallest mole value (2.38mol)
C = 2.38mol ÷ 2.38 = 1
H = 4.80mol ÷ 2.38 = 2.01
N = 4.76mol ÷ 2.38 = 2
The empirical ratio of C, H and N is therefore 1:2:2. Hence, the empirical formula is CH2N2
To calculate the molecular formula;
(CH2N2)n = 42.04 g/mol
{12 + 1(2) + 14(2)}n = 42.04
{12 + 2 + 28}n = 42.04
{42}n = 42.04
n = 42.04/42
n = 1.00009
Since n = 1, molecular formula is CH2N2
Answer:
A precipitate will be formed
Explanation:
The Ksp equilibrium of Fe(OH)₃ is:
Fe(OH)₃ (s) ⇄ Fe³⁺(aq)+ 3OH⁻(aq)
And its expression is:
Ksp = 4x10⁻³⁸ = [Fe³⁺] [OH⁻]³
<em>Where the concentrations are concentrations in molarity in equilibrium,</em>
We can write Q as:
Q = [Fe³⁺] [OH⁻]³
<em>Where [] are actual concentrations in molarity of each specie.</em>
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When Q>= Ksp; a precipitate is formed,
When Q< Ksp no precipitate is produced:
[OH⁻] = [NaOH] = 1.0x10⁻⁴M
[Fe²⁺] = 2.50x10⁻²g * (1mol / 179.85g) / 0.100L = 1.39x10⁻³M
<em>179.85g/mol is molar mass of Fe(NO₃)₂ and the volume of the solution is 0.100L = 100mL</em>
<em />
Q = [Fe³⁺] [OH⁻]³
Q = [ 1.39x10⁻³] [ 1.0x10⁻⁴]³
Q = 3.8x10⁻¹⁵
As Q >> Ksp; A precipitate will be formed
Answer:
1s22s22p3.
Explanation:
Electronic configuration of a neutral atom is 1s22s22p3.
Please see the image attached
Neutral atom of nitrogen will have equal number of proton and electron i.e equal to 7. 7 electron of the nitrogen are placed into the s and p orbitals in the ground state.
