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:
0.0164 g
Explanation:
Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.
Ag⁺(aq) + 1 e⁻ → Ag(s)
We can establish the following relations.
- 1 A = 1 C/s
- The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
- 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
- The molar mass of silver is 107.87 g/mol
The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

Answer:
The other signal that would indicate the presence of a C= C bond appears close to 3100
.
Explanation:
Bands that appear above 3000
are often unsaturation diagnoses suggest. The band at 3000-
3100
is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.
Answer:
The enthalpy of the reaction is –184.6 kJ, and the reaction is exothermic.
Explanation:
<u>Answer:</u> This illustrates law of conservation of mass.
<u>Explanation:</u>
Dalton's theory is based on mainly two laws which are law of conservation of mass and law of constant proportion.
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
The chemical equation for the decomposition of calcium carbonate follows:

We are given:
Mass of calcium carbonate = 100 grams
Mass of calcium oxide = 56 grams
Mass of carbon dioxide = 44 grams
Total mass on reactant side = 100 g
Total mass on product side = 56 + 44 = 100 g
As, the total mass on reactant side is equal to the total mass on product side.
Thus, this illustrates law of conservation of mass.