The problem can be solved using the following formula:
ΔTb = i Kb <em>m</em>
i = moles particles/moles solute
Kb = 0.512 °C/m
m = molality = moles solute/kg solvent
First we can solve for the molality of the solution:
75.0 g ZnCl₂ / 136.286 g/mol = 0.550 mol ZnCl₂
m = 0.550 mol/0.375 kg
m = 1.468 mol/kg
We can now solve for the change in temperature of the boiling point:
ΔTb = i Kb m
ΔTb = (3 mol particles/1 mol ZnCl₂) (0.512 °C/m) (1.468 m)
ΔTb = 2.25 °C
The boiling point of a solution is the initial boiling point plus the change in boiling point:
BP = 100 °C + 2.25 °C
BP = 102.25 °C
The solution will have a boiling point of 102.25 °C.
Answer:
CaO
Explanation:
CaO is the only compound that is a non-metal and a non-metal. The rest of the compounds are ionic, or metal and non-metal.
The molar mass of the imaginary compound Z(AX₃)₂ is the sum of the molar mass of Z, A and X.
<h3>How do we calculate molar mass?</h3>
Molar mass of any compound will be calculated by adding the mass of each atoms present in that compound.
Given compound is Z(AX₃)₂, molar mass of the given compound will be calculated as:
Molar mass of Z(AX₃)₂ = Molar mass of Z + molar mass of 2(A) + molar mass of 6(X)
Hence molar mass of Z(AX₃)₂ is the sum of the masses of all atoms.
To know more about molar mass, visit the below link:
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Explanation:
to calculate theoratical yield...first we must calculate how much Sucrose moles we have..in order to to do so...we must calculate molar mass of sucrose.
molar mass of sucrose is 342.3 g/mol.
now we can calculate how many sucrose moles we have by dividing the mass with molar mass of sucrose
735g/(342.3g/mol)=2.147mol
theoratically...according to stoichiometry..every 1 mole of sucrose yields 4 moles of ethanol...so...
2.147mole yield 2.147*4mol=8.588mol
now we must calculate weight of that much ethanol..molar mass of ethanol is 46.07 g/mol...
so we can multiple moles by molar mass to obtain the weight 8.588mol*46.07g/mol=395.649g
but we only obtained 310.5g...so percentage we have is

78.47%
if there is trouble with molar mases I used....use what you calculated...
If my explanation is good enough...please mark it as brainliest.thanks