<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:
NaOH(aq) is a Base.
Explanation:
Those substances which give or release 
ions in aqueous solution are called as the Arrhenius Bases.
In the aqueous solution, NaOH dissociates as follows -
↔ 
If it reacts with a strong acid HCl, the chemical equation for this reaction will be as follows -
The solution for the question above is:
C = 0.270
<span>V = 0.0275L </span>
<span>n = ? </span>
<span>Use the molar formula which is: C = n/V </span>
<span>Re-arrange it to: n = CV </span>
<span>n = (0.270)*(0.0275) </span>
<span>n = 0.007425 mols </span>
<span>(more precise) n = 7.425 x 10^-3 mols
</span>
7.425 x 10^-3 mols is the answer.
Answer:
It's the third option.
Explanation:
In order for the chemical equation to be correctly it needs the same number of atoms of each element on both sides of the equal sign
D) if stuff is changing then the reaction is hardly in equilibrium is it? Everything is just chilling at equilibrium so there would be constant concentration
