<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
It’s the second answer. “it gets used for metabolic process & released as heat” i think.
Answer:
2KClO3 —> 2KCl + 3O2
The coefficients are 2, 2, 3
Explanation:
From the question given above, we obtained the following equation:
KClO3 —> 2KCl + 3O2
The above equation can be balance as follow:
There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:
2KClO3 —> 2KCl + 3O2
Now, the equation is balanced.
Thus, the coefficients are 2, 2, 3
they could be different types of minerals, so one reacted to the energy differently than the other one. and pls answer my latest questions its due by 11:59 pm TONIGHT or i get kicked out of school.. pls help..
It forms <span>calcium phosphate and potassium nitrate
</span>2 K3PO4 + 3Ca(NO3)2 --> Ca3(PO4)2 + 6KNO3