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Molecular Weight: 262.86 g/mol
Volume in liters : 1,000 mL => 1,000 / 1000 => 1 L
Number of moles of solution :
number of moles = volume x molarity
number of moles = 1 x 0.3 => 0.3 moles
Therefore:
1 mole Mg₃(PO₄)₂ ------------------- 262.86 g
0.3 moles ----------------------------- mass of Mg₃(PO₄)₂ ??
mass of Mg₃(PO₄)₂ = 0.3 x 262.86 / 1
mass of Mg(PO₄)₂ => 78.858 g
Hope this helps!
Answer:
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Answer:
0.02moles
Explanation:
To answer this question, the general gas law equation is used. The General gas law is:
Pv = nRT
Where; P = standard atmospheric pressure (1 atm)
V = volume (L)
n = number of moles
R = Gas law constant
T = Temperature
For this question; volume = 1.00L, atmospheric pressure (P) = 1 atm, R = 0.0821 L-atm / mol K, T = 600K, n = ?
Therefore; Pv = nRT
n = PV/RT
n = 1 × 1/ 0.0821 × 600
n = 1/49.26
n = 0.0203moles
Hence, there are 0.02 moles of gas.