Answer:
Both have the same amount of particles.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles.
This implies that 1 mole of Hydrogen contains 6.02×10²³ particles. Also, 1 mole of oxygen contains 6.02×10²³ particles.
Thus, 1 mole of Hydrogen and 1 mole of oxygen contains the same number of particles.
The net equations are obtained from the double displacement of the cations and anions, then balance.
NH3(aq) + HC2H3O2 (aq) = NH4+(aq) + C2H3O2-(aq<span>)
</span><span>H+(aq) + C2H3O2-(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq)</span><span>
</span><span>2NaOH(aq) + H2SO4 (aq) = Na2SO4 (s)+ 2H2O (aq)
</span>H2S (aq) + Ba(OH)2 (aq) = BaS (s)+ 2H2O (aq)
First convert the kg to g ----- 0.03kg = 30g
Then divide the mass by the volume ----- 30g ÷ 25mL = 1.2
The density is 1.2g/mL<span />
Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
