Answer is: mass of the ore is 8.54kg.<span>
</span>ω(Ca₃(PO₄)₂ - calcium phosphate) = 58.6% ÷ 100% = 0.586.
m(P) = 1.00 kg · 1000 g/kg.
m(P) = 1000 g.
In one molecule of calcium phosphate there are two phosphorus atoms:
M(Ca₃(PO₄)₂) = 310.18 g/mol.
M(P) = 30.97 g/mol.
For one kilogram of phosphorus, we need:
M(Ca₃(PO₄)₂) : 2M(P) = m(Ca₃(PO₄)₂) : m(P).
310.18 g/mol : 61.94 g/mol = m(Ca₃(PO₄)₂) : 1000 g.
m(Ca₃(PO₄)₂) = 5007.75 g ÷ 1000 g/kg = 5.007 kg.
Mass of ore find from proportion:
m(Ca₃(PO₄)₂) : m(ore) = 56% : 100%.
m(ore) = 100% · 5.007 kg ÷ 58.6%.
m(ore) = 8.54kg.
Answer:
you can solve the rest of the equation. I only reduced it to that much to show you how to derive it
Answer:
joules
Explanation:
it is the measurement of energy
Empirical formula is the simplest ratio of components making up a compound.
The percentage composition of each element has been given
therefore the mass present of each element in 100 g of compound is
B N H
mass 40.28 g 52.20 g 7.53 g
number of moles
40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol
= 3.662 mol = 3.729 mol = 7.53 mol
divide the number of moles by the least number of moles, that is 3.662
3.662 / 3.662 3.729 / 3.662 7.53 / 3.662
= 1.000 = 1.018 = 2.056
the ratio of the elements after rounding off to the nearest whole number is
B : N : H = 1 : 1 : 2
therefore empirical formula for the compound is B₁N₁H₂
that can be written as BNH₂
