81.875 g of magnesium phosphate Mg₂(PO₄)₃
Explanation:
The chemical formula of magnesium phosphate is Mg₃(PO₄)₂.
molar mass of Mg₃(PO₄)₂ = atomic weight of Mg × 3 + atomic weight of P × 3 + atomic weight of O × 3 × 4
molar mass of Mg₃(PO₄)₂ = 24 × 3 + 31 × 2 + 16 × 2 × 4 = 262 g/mole
Knowing the molar mass of Mg₃(PO₄)₂, we devise the following reasoning:
if in 262 g of Mg₂(PO₄)₃ there are 48 g of Mg
then in X g of Mg₂(PO₄)₃ there are 15 g of Mg
X = (262 × 15) / 48 = 81.875 g of Mg₂(PO₄)₃
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Answer:
c) 9.03 x 10^23
Explanation:
find the molar mass of Al
Al is 27.0 grams
Then use that, to find the number of moles in Aluminum.
Then use Avogadro's number which is 6.02 * 10^23
After that, write all of that down with dimensional analysis.
40.5 g * 1 mol/ 27.0 g of Al * 6.02 x 10^23 / 1mol
As your final answer, you will get 9.03 * 10^23 atoms with sig figs.
Hope it helped!
Answer:
The correct answer is 16 gram per mole.
Explanation:
Let A be the gas helium, and B be the unknown gas. It is clearly mentioned in the question that the effusion rate of helium gas is two times more than that of gas B. The molar mass of helium is 4.0 gram per mole. To solve the problem, Grahm's law is used, that is,
Rate of effusion A/rate of effusion B = √ (Molar mass B/Molar mass of A
2.0 = √Molar mass of B/4.0 gram per mole.
Now squaring both the sides we get,
4.0 = Molar mass of B / 4.0
The molar mass of B = 16 gram per mole.
It would be two because there’s 1 Carbon and 2 oxygen
Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
