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:
V2 = 894.4mL
Explanation:
P1= 124.1, V1= 578mL, P2 = 80.2kPa, V2= ?
Applying Boyle's law
P1V1 = P2V2
Substitute and simplify
124.1*578=80.2*V2
V2= 894.4mL
Answer:
In the chemical industry.
Answer:
46.3g H2O
Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)
then use factor label method to solve
82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
