I think the Ksp for Calcium Carbonate is around 5×10⁻⁹
(I don't know if this is the Ksp value that you use because I read somewhere that this value can vary. You should probably check with your teacher with what Ksp value they want you to use)
the equation for the dissociation CaCO₃ in water is CaCO₃(s)⇄Ca²⁺(aq)+CO₃²⁻(aq) which means that the concentration of Ca²⁺ is equal to the concentration of CO₃²⁻ in solution. For every molecule of CaCO₃ that dissolves, one atom of Ca²⁺ and one molecule of CO₃²⁻ is put into solution which is why the concentrations are equal in solution.
Since Ksp=[Ca²⁺][CO₃²⁻] and we know that [Ca²⁺]=[CO₃²⁻] we can rewrite the equation as Ksp=x² since if you say that [Ca²⁺]=[CO₃²⁻] when you multiply them together you get the concentration squared (I am calling the concentration x for right now).
when solving for x:
5×10⁻⁹=x²
x=0.0000707
Therefore [Ca²⁺]=[CO₃²⁻]=0.0000707mol/L which also shows how much calcium carbonate is dissolved per liter of water since the amount of Ca²⁺ and CO₃²⁻ in solution came from the calcium in a 1 to 1 molar ratio as shown in the equation (the value we found for x is the molar solubility of calcium carbonate).
Using the fact that the molar mass of calcium carbonate is 100.09g/mol you can use dimensional analysis as fallows:
(0.0000707mol/L)(100.09g/mol)=0.007077g/L
That means that there is 0.007077g of Calcium carbonate that can precipitate out of 1L of water.
since the question is asking for how much water needs to be evaporated to precipitate 100mg (0.1g) of Calcium you have to do the fallowing calculation:
(0.1g)/(0.007077g/L)=14.13L of water.
14.13L of water needs to evaporate in order to precipitate out 100mg of calcium carbonate
These types of questions can get long and confusing so I bolded parts that were important to try to guide you through it more easily.
I hope this helps. Let me know if anything is unclear.
Answer:
Oxidative phosphorylation proceeds with the formation of energy laden molecules i.e; carbondioxide and water.
Therefore, Total CO₂ production is directly related to VCO₂ = R x VO₂
where, R is the respiratory quotient varing among 0.7 to 1.0 according to the energy intake (ATP) ie 0.25 of the total diet consumed .
VO₂ is, as mentioned above arterial venous oxygen difference = 6.2ml/dl
therefore, VCO₂ = 0.25 x 6.2
= 1.55 ml/dl
ie; VO₂ : VCO₂ = 6.2 : 1.55.
Explanation:
First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo
We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g
We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O
To convert mass to moles, we use the molar mass of each element.
1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O
0.0131 mol is the smallest number of moles. We divide each mole value by this number:
0.0131 mol Mo / 0.0131 = 1
0.0200 mol O / 0.0131 = 1.53
Multiplying these results by 2 to get the lowest whole number ratio,
0.0131 mol Mo / 0.0131 = 1 * 2 = 2
0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.
When ΔG° is the change in Gibbs free energy
So according to ΔG° formula:
ΔG° = - R*T*(㏑K)
here when K = [NH3]^2/[N2][H2]^3 = Kc
and Kc = 9
and when T is the temperature in Kelvin = 350 + 273 = 623 K
and R is the universal gas constant = 8.314 1/mol.K
So by substitution in ΔG° formula:
∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)
= - 4536
