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>
<em></em>
Mass percentage of a solution is the amount of solute present in 100 g of the solution.
Given data:
Mass of solute H2SO4 = 571.3 g
Volume of the solution = 1 lit = 1000 ml
Density of solution = 1.329 g/cm3 = 1.329 g/ml
Calculations:
Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g
Therefore we have:
571.3 g of H2SO4 in 1329 g of the solution
Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987
Mass percentage of H2SO4 (%w/w) is 42.99 %
Answer:
C
Explanation:
Pressure it says it in the name its the force of colliding particles
<u>Answer:</u> The number of moles of weak acid is 
moles.
<u>Explanation:</u>
To calculate the moles of KOH, we use the equation:
We are given:
Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)
Molarity of the solution = 0.0969 moles/ L
Putting values in above equation, we get:
The chemical reaction of weak monoprotic acid and KOH follows the equation:
By Stoichiometry of the reaction:
1 mole of KOH reacts with 1 mole of weak monoprotic acid.
So, 
of KOH will react with = 
of weak monoprotic acid.
Hence, the number of moles of weak acid is moles.
