Answer:
6.4 × 10^-10 M
Explanation:
The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).
CaF2 will dissociate as follows:
CaF2 ⇌Ca2+ + 2F-
1 mole of Calcium ion (x)
2 moles of fluorine ion (2x)
NaF will also dissociate as follows:
NaF ⇌ Na+ + F-
Where Na+ = 0.25M
F- = 0.25M
The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M
Ksp = {Ca2+}{F-}^2
Ksp = {x}{0.25}^2
4.0 × 10^-11 = 0.25^2 × x
4.0 × 10^-11 = 0.0625x
x = 4.0 × 10^-11 ÷ 6.25 × 10^-2
x = 4/6.25 × 10^ (-11+2)
x = 0.64 × 10^-9
x = 6.4 × 10^-10
Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M
Answer:
116.3 grCO2
Explanation:
1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side
C6H6 +15/2 O2⟶ 6CO2 +3 H2O
2nd - we calculate the limiting reagent
39.2gr C6H6*(240grO2/78grC6H6)=120 grO2
we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent
3rd - we use the limiting reagent to calculate the amount of CO2 in grams
105.7grO2*(264grCO2/240grO2)=116.3 grCO2
In comparison with liquids and gases, solids are more dense. The answer is letter B. <span>The
solid has a more definite shape and volume. The particles are locked into
place. It cannot be further compressed due to the bond that exists between the
molecules. The kinetic energy of the molecules is close to none because the
molecules are so close and so compact with each other. </span>
