When the value of Ksp = 3.83 x 10^-11 (should be given - missing in your Q)
So, according to the balanced equation of the reaction:
and by using ICE table:
Ag2CrO4(s) → 2Ag+ (Aq) + CrO4^2-(aq)
initial 0 0
change +2X +X
Equ 2X X
∴ Ksp = [Ag+]^2[CrO42-]
so by substitution:
∴ 3.83 x 10^-11 = (2X)^2* X
3.83 x 10^-11 = 4 X^3
∴X = 2.1 x 10^-4
∴[CrO42-] = X = 2.1 x 10^-4 M
[Ag+] = 2X = 2 * (2.1 x 10^-4)
= 4.2 x 10^-4 M
when we comparing with the actual concentration of [Ag+] and [CrO42-]
when moles Ag+ = molarity * volume
= 0.004 m * 0.005L
= 2 x 10^-5 moles
[Ag+] = moles / total volume
= 2 x 10^-5 / 0.01L
= 0.002 M
moles CrO42- = molarity * volume
= 0.0024 m * 0.005 L
= 1.2 x 10^-5 mol
∴[CrO42-] = moles / total volume
= (1.2 x 10^-5)mol / 0.01 L
= 0.0012 M
by comparing this values with the max concentration that is saturation in the solution
and when the 2 values of ions concentration are >>> than the max values o the concentrations that are will be saturated.
∴ the excess will precipitate out
<u>Answer:</u> The molarity of solution is 1.08 M
<u>Explanation:</u>
We are given:
(m/m) of phenol = 1.40 %
This means that 1.40 g of phenol is present in 100 g of solution.
To calculate volume of solution, we use the equation:

Density of solution = 0.9956 g/mL
Mass of solution = 100 g
Putting values in above equation, we get:

To calculate the molarity of solution, we use the equation:

We are given:
Mass of solute (phenol) = 1.40 g
Molar mass of phenol = 94.11 g/mol
Volume of solution = 100.442 mL
Putting values in above equation, we get:

Hence, the molarity of solution is 0.15 M
<span>this is a limiting reagent problem.
first, balance the equation
4Na+ O2 ---> 2Na2O
use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make.
start with Na and go to grams of Na2O
55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O
do the same with O2
64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O
now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops.
So, the mass of sodium oxide is
75.5 g</span>
It’s the last one icebergs shrink as global temp rises