Answer:
A gas made up of homonuclear diatomic molecules escapes through a pinhole .533 times as fast as Ne gas. Write the chemical formula of the gas.
Molar mass of Ne gas is 39.1 g/mol
.533= sqrt(39.1/x)
Sq (0.533) = 20.18/x
0.284 = 20.18/x
x = 71.034 g/mol
Where the chlorine has a molar mass of = 35.5g and molar mass of Cl2 gas = 2 × 35.5 = 71 g/mol
Explanation:
Graham's law: Rate1/Rate2 = sqrt(M2/M1) where M is the molar mass
Thus we have (Rate of x)/(Rate of Ne) = sqrt((Molar mass of Ne)/(Molar mass of x))
From the question (Rate of x)/(Rate of Ne) =0.533
and sqrt((Molar mass of Ne)/(Molar mass of x)) = sqrt(39.1/(Molar mass of x))
.533/1= sqrt(39.1/x)
Sq (0.533) = 20.18/x
0.284 = 20.18/x
x = 71.034 g/mol
Hence a compatible gas is chlorine Cl2 with molar mass of
71 g/mol
The third class lever s<span>have </span>the effort<span> placed amongst </span>load<span> and the fulcrum.</span>
As the atomic mass of iron is 55.847u
We say that it is the mass of one mole of iron.
By formula it can be find by
No.of mole=mass in g/molar mass
Mass in gram = no.of mole x molar mass
No.of mole =1
Molar mass = 55.847g
Mass in gram = 1x55.847
= 55.847g of Fe
Answer:
See explanation
Explanation:
In a chemical reaction, a particular reaction path may be favoured due to the fact that it is energetically more favourable(lower energy sigma complex is formed).
The more the resonance structures produced in a particular reaction pathway, the more energetically favourable it is.
In the chlorination of bromobenzene, ortho attack and para attack are preferred because each of these pathways involves a sigma complex with__4________resonance structures. Attack at the meta position involves formation of a sigma complex with only____3_______ resonance structures. The reaction will proceed more rapidly via the_____lower_______ energy sigma complex, so attack takes place at the ortho and para positions in preference to the meta position.
Answer:
Volume of N₂ = 14.76 L
Volume of H₂ = 29.52 L
Explanation:
Given data:
Mass of N₂H₄ formed = 28.5 g
Pressure = 1.50 atm
Temperature = 30°C (30+273 = 303 k)
Volume of N₂ and H₂ needed = ?
Solution:
Chemical equation:
N₂ + 2H₂ → N₂H₄
Number of moles of N₂H₄ formed = mass/ molar mass
Number of moles of N₂H₄ formed = 28.5 g/ 32 g/mol
Number of moles of N₂H₄ formed = 0.89 mol
Now we will compare the moles of N₂H₄ with N₂ and H₂ form balance chemical equation.
N₂H₄ : N₂
1 : 1
0.89 : 0.89
N₂H₄ : H₂
1 : 2
0.89 : 2×0.89 = 1.78 mol
Volume of H₂:
PV = nRT
1.50 atm × V = 1.78 mol × 0.0821 atm.L/mol.K × 303 K
V = 44.28atm.L /1.50 atm
V = 29.52 L
Volume of N₂:
PV = nRT
1.50 atm × V = 0.89 mol × 0.0821 atm.L/mol.K × 303 K
V = 22.14 atm.L /1.50 atm
V = 14.76 L
