Calculate the volume of a 2.30 mol/L solution of Na2CO3 that contains 0.325 mol of solute.
1 answer:
Answer:
141 mL is the volume of a 2.30 mol/L solution of Na₂CO₃ that contains 0.325 mol of solute.
Explanation:
Molarity is a measure of concentration of a solution that indicates the amount of moles of solute that appear dissolved in each liter of the mixture.
So 2.30 mol / L indicates that 2.30 moles dissolve in 1 L of solution. You want to know in what volume is contained 0.325 mol of solute. For this you apply the following rule of three: if 2.30 moles of solute are present in 1 L of solution, 0.325 moles in how much volume is it?
volume= 0.141 L
Being 1L=1000 mL, then 0.141 L=141 mL
<u><em>141 mL is the volume of a 2.30 mol/L solution of Na₂CO₃ that contains 0.325 mol of solute.</em></u>
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Answer:
Oxide of M is and sulfate of
Explanation:
0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.
Let moles of hydrogen gas be n.
Temperature of the gas ,T= 17°C =290 K
Pressure of the gas ,P= 741 mmHg= 0.9633 atm
Volume occupied by gas , V = 0.303 L
Using an ideal gas equation:
Moles of hydrogen gas produced = 0.01225 mol
Moles of metal =
So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.
x = 2.9 ≈ 3
Formulas for the oxide and sulfate of M will be:
Oxide of M is and sulfate of
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