It is important to use low flame when evaporating water from a recovered filtrate because then the water and filtrate will not spatter and the filtrate can also be recovered after evaporating water.
If flame is not low then water as well as got spatter so it is important to use low flame so that the water and filtrate will not spatter.
Both involve chemical bonds breaking
<span>to find the molarity of an unknown acid or base</span>
Answer:
The final molarity of acetate anion in the solution is 0.0046 moles
Explanation:
The balanced equation is
Cu(C₂H₃O₂)₂ + Na₂CrO₄ = CuCrO₄ + 2Na(C₂H₃O₂)
Therefore one mole of Cu(C₂H₃O₂)₂ react with one mole of Na₂CrO₄ to form one mole of CuCrO₄ and two moles of Na(C₂H₃O₂)
Mass of copper (II) acetate present = 0.708 g
Volume of aqueous sodium present = 50 mL
Molarity of sodium chromate = 46.0 mM
Therefore
Number of moles of sodium chromate present = (50 mL/1000)×46/1000 = 0.0023 M
Number of moles of copper (II) acetate present = 181.63 g/mol
number of moles of copper (II) acetate present = (0.708 g/181.63 g/mol) =0.0039 moles
Therefore 0.0039 moles of Cu(C₂H₃O₂)₂ × (2 moles of Na(C₂H₃O₂))/1 Cu(C₂H₃O₂)₂) = 0.00779 moles of Na(C₂H₃O₂)
also 0.0023 moles of Na₂CrO₄ × (2 moles of Na(C₂H₃O₂))/1 Na₂CrO₄) = 0.0046 moles of Na(C₂H₃O₂)
Therefore the Na₂CrO₄ is the limiting reactant and 0.0046 moles of Na(C₂H₃O₂) or acetate anion is formed
Answer:
138mL are required to make the solution
Explanation:
A solution of 0.270M contains 0.270 moles of solute (NaCl in this case), per liter of solution.
To solve this problem we need to determine the moles of NaCl in 2.18g. With these moles and molarity we can find the volume required to prepare the solution:
<em>Moles NaCl -Molar mass: 58.44g/mol-:</em>
2.18g * (1mol / 58.44g) = 0.0373moles NaCl
<em>Volume of solution:</em>
0.0373moles NaCl * (1L / 0.270moles) = 0.138L
<h3>138mL are required to make the solution</h3>
