Explanation:
Amount of water required in each case:
(a)The mass% of the solution is:9.95
Mass of solute that is urea is 6.80 g
To determine the mass of solvent water use the formula:
Hence the mass of solvent = mass of solution - the mass of solute
=68.3 g - 6.80g
=61.5 g
Hence, the answer is mass of solvent water required is 61.5 g.
(b) Given mass%=1.70
mass of solute MgBr2 = 29.3 g
The mass of solvent water required can be calculated as shown below:
The mass of the solution is 1720 g.
Mass of solvent water = mass of solution - mass of solute
=1720 g - 29.3 g
=1690.7 g
Answer: The mass of water required is 1690.7 g.
Answer : The equilibrium concentration of CO in the reaction is, 
Explanation :
The given chemical reaction is:
The expression for equilibrium constant is:
![K_c=\frac{[COCl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCOCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
As we are given:
Concentration of 
at equilibrium = Concentration of 
So,
![K_c=\frac{[Cl_2]}{[CO][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCl_2%5D%7D%7B%5BCO%5D%5BCl_2%5D%7D)
![K_c=\frac{1}{[CO]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B1%7D%7B%5BCO%5D%7D)
![1.2\times 10^3=\frac{1}{[CO]}](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E3%3D%5Cfrac%7B1%7D%7B%5BCO%5D%7D)
![[CO]=8.3\times 10^{-4}M](https://tex.z-dn.net/?f=%5BCO%5D%3D8.3%5Ctimes%2010%5E%7B-4%7DM)
Therefore, the equilibrium concentration of CO in the reaction is,
What data are you showing ????
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
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
