Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
Answer : The correct options are,
(B) 
(C) 
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
or,
The relation between the pressure and volume of two gases are:
where,
= initial pressure of gas
= final pressure of gas
= initial volume of gas
= final volume of gas
Answer:
Amino >Methoxy > Acetamido
Explanation:
Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.
The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.
The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.
Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.
Therefore, decreasing order will be as follows:
Amino >Methoxy > Acetamido
Answer:
Mass released = 8.6 g
Explanation:
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
