The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
The heat of combustion for methanol is 727 kj/mol
<em><u>calculation</u></em>
calculate the moles of methanol (CH3OH)
moles = mass/molar mass
molar mass of methanol = 12 +( 1 x3) +16 + 1= 32 g /mol
moles is therefore= 64.0 g / 32 g/mol = 2 moles
Heat of combustion is therefore = 1454 Kj / 2 moles = 727 Kj/mol
Answer:
NaI > Na2SO4 > Co Br3
meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.
Explanation:
The freezing point depression is a colligative property.
That means that it depends on the number of solute particles dissolved.
The formula to calculate the freezing point depression of a solution of a non volatile solute is:
ΔTf = i * Kf * m
Where kf is a constant, m is the molality and i is the van't Hoff factor.
Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.
The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.
As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:
NH4 I → NH4(+) + I(-) => 2 ions
Co Br3 → Co(+) + 3 Br(-) => 4 ions
Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.
So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.
Answer:
Gamma
Explanation:
I'm not sure how to do it without calculations but:
E=hv
7*10^7 J/mol=6.626*10^34 Js * v
v=1*10^41
Gamma rays.
More here: https://www.hasd.org/faculty/AndrewSchweitzer/spectroscopy.pdf
Answer:
First one is 5.0 M ammonia and the Second one ?
Explanation:
