Answer:
B. A precipitate will form since Q > Ksp for calcium oxalate
Explanation:
Ksp of CaC₂O₄ is:
CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻
Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:
Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹
In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.
Replacing in Ksp formula:
[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.
If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.
Thus, right answer is:
<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>
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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.
The density of an object is defined as its mass divided by its volume. Mathematically, density = Mass / Volume. The unit of density is kilogram per cubic meter, kg / m^3 or g /cm^3.
For the question given above: the
Mass = 200.0 g
Volume = 100.0 cm^3
Therefore, Density = Mass / Volume = 200 / 100 = 2
Thus, the density of the object is 2 g /cm^3.
Answer:
Just add what they equal to hope this helps :)
Explanation: