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:
Moles of 
= 0.00026 moles
Explanation:
Mass of = 0.2589 g
Molar mass of = 100.0869 g/mol
The formula for the calculation of moles is shown below:
Thus,
Also,
Volume = 250 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 250×10⁻³ L = 0.250 L
So,
Molarity of the sample = 0.0104 M
Considering:
Volume = 25.00 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 25.00×10⁻³ L
Thus, moles of :
Moles of = 0.00026 moles
<u>Superior</u> means anything which is higher in rank , status or quality among similar other things or people.
<u>Posterior</u> means something which is at back.
<u>Inferior</u> means anything which is lower in rank , status or quality among similar other things or people.
<u>Anterior</u> means something which is nearer to the front.
∴ Pair of opposite terms is: Superior/inferior
The equivalence point is when the concentration of H⁺ in solution is equal to the concentration of OH⁻ in solution. Since H⁺ and OH⁻ react with each other to make water (H⁺(aq)+OH⁻(aq)→H₂O(l)) the pH at the equivalence point is 7 due to everything being neutralized. (The equivalence point only has a pH of 7 when a strong acid is being titrated with a strong base).
I hope this helps. Let me know if anything is unclear.
