<h3>
Answer:</h3>
2.624 g
<h3>
Explanation:</h3>
The equation for the reaction is given as;
- CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
- Volume of CuSO₄ as 46.0 mL;
- Molarity of CuSO₄ as 0.584 M
We are required to calculate the mass of Cu(OH)₂ precipitated
- We are going to use the following steps;
<h3>Step 1: Calculate the number of moles of CuSO₄ used</h3>
Molarity = Number of moles ÷ Volume
To get the number of moles;
Moles = Molarity × volume
= 0.584 M × 0.046 L
= 0.0269 moles
<h3>
Step 2: Calculate the number of moles of Cu(OH)₂ produced </h3>
- From the equation 1 mole of CuSO₄ reacts to give out 1 mole of Cu(OH)₂
- Therefore; Mole ratio of CuSO₄ to Cu(OH)₂ is 1 : 1.
Thus, Moles of CuSO₄ = Moles of Cu(OH)₂
Hence, moles of Cu(OH)₂ = 0.0269 moles
<h3>
Step 3: Calculate the mass of Cu(OH)₂</h3>
To get mass we multiply the number of moles with the molar mass.
Mass = Moles × Molar mass
Molar mass of Cu(OH)₂ is 97.561 g/mol
Therefore;
Mass of Cu(OH)₂ = 0.0269 moles × 97.561 g/mol
= 2.624 g
Thus, the mass of Cu(OH)₂ that will precipitate is 2.624 g
Answer:
Moderate reaction means those reactions which proceed with a measurable rates at the normal room temperature.
Explanation:
Filtration is a separation technique in which solid particles suspended in liquid medium are separated by allowing the mixture through the pores of the filter paper. By this solid particles get collect on filter paper and liquid drains out from the pores of the filter paper.
The chronological order for given steps will be:
- Weigh and fold the filter paper.
- Place the filter paper in the funnel, then place the funnel in the Erlenmeyer flask.
- Allow the solid/liquid mixture to drain through the filter.
- Use water to rinse the filter paper containing the mixture.
- Weigh the dried filter paper and copper.
Answer:
The initial temperature of the metal is 84.149 °C.
Explanation:
The heat lost by the metal will be equivalent to the heat gain by the water.
- (msΔT)metal = (msΔT)water
-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C
-ΔT = 836.29/12.51 °C
-ΔT = 66.89 °C
-(T final - T initial) = 66.89 °C
T initial = 66.89 °C + T final
T initial = 66.89 °C + 17.3 °C
T initial = 84.149 °C.