Answer:
there is 2% of hydrogen and 98% of nitrogen (mass percent)
Explanation:
assuming ideal gas behaviour
P*V=n*R*T
n= P*V/(R*T)
where P= pressure=1.02 atm , V=volume=7.47 L , T=absolute temperature= 296 K and R= ideal gas constant = 0.082 atm*L/(mole*K)
thus
n= P*V/(R*T) = 1.02 atm*7.47 L/( 296 K * 0.082 atm*L/(mole*K)) = 0.314 moles
since the number of moles is related with the mass m through the molecular weight M
n=m/M
thus denoting 1 as hydrogen and 2 as nitrogen
m₁+m₂ = mt (total mass)
m₁/M₁+m₂/M₂ = n
dividing one equation by the other and denoting mass fraction w₁= m₁/mt , w₂= m₂/mt , w₂= 1- w₁
w₁/M₁+w₂/M₂ = n/mt
w₁/M₁+(1-w₁) /M₂ = n/mt
w₁*(1/M₁- 1/M₂) + 1/M₂ = n/mt
w₁= (n/mt- 1/M₂) /(1/M₁- 1/M₂)
replacing values
w₁= (n/mt- 1/M₂) /(1/M₁- 1/M₂) = (0.314 moles/3.48 g - 1/(14 g/mole)) /(1/(1 g/mole)-1/(14 g/mole))= 0.02 (%)
and w₂= 1-w₁= 0.98 (98%)
thus there is 2% of hydrogen and 98% of nitrogen
Answer:
1.63425 × 10^- 18 Joules.
Explanation:
We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.
We will be using the formula in the equation (1) below;
Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).
Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.
Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;
∆E= - R(H) × [1/nf^2 - 1/ni^2 ].
Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.
Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].
= -2.179 × 10^-18 Joules × (0.25 - 1).
= - 2.179 × 10^-18 × (- 0.75).
= 1.63425 × 10^- 18 Joules.
Answer:
It is double displacement reaction
2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
