Answer: Thus 0.724 mol of
are needed to obtain 18.6 g of 
Explanation:
To calculate the moles :

According to stoichiometry :
2 moles of
are produced by = 1 mole of 
Thus 1.09 moles of
will be produced by =
of 
But as yield of reaction is 75.6 %, the amount of
needed is =
Thus 0.724 mol of
are needed to obtain 18.6 g of 
Answer:
1-46
2-18
Explanation:
c=12 H=1 O=16
ethanol (12×2)+(6×1)+(16)=46
water (2×1)+(16)=18
Answer:
5.37 × 10⁻⁴ mol/L
Explanation:
<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.00154 mol/L
- Initial volume (V₁): 230. mL
- Final concentration (C₂): ?
- Final volume (V₂): 660. mL
Step 2: Calculate the concentration of the final solution
We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁ / V₂
C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L
Answer:

Explanation:
According to the boiling point elevation law described by the equation
, the increase in boiling point is directly proportional to the van 't Hoff factor.
The van 't Hoff factor for nonelectrolytes is 1, while for ionic substances, it is equal to the number of moles of ions produced when 1 mole of salt dissolves.
would produce 2 moles of ions per 1 mole of dissolved substance, sodium and bromide ions.
is insoluble in water, so it would barely dissociate and wouldn't practically change the boiling point.
would dissociate into 3 moles of ions per 1 mole of substance, two potassium cations and one sulfide anion.
is a gas, it would form some amount of carbonic acid when dissolved, however, carbonic acid is molecular and would yield i value of i = 1.
Therefore, potassium sulfide would raise a liquid's boiling point the most if all concentrations are equal.