2 Na₂CO₃ + 2 CuSO₄ + H₂O → CuCO₃.Cu(OH)₂ + 2 Na₂SO₄ + CO₂
Malachite molar mass = 221.1 g / mol
So 2 moles CuSO₄ produce 1 mole of malachite
so 1.5 mole CuSO₄ produce (0.75) mole malachite
Mass of malachite = 0.75 mole * 221.1 g/ mol = 165.83 g
Answer:
V = 10.3 L
Explanation:
Given data:
Mass of methane = 6.40 g
Volume of CO₂ produced = ?
Temperature = 35°C (35+273 = 308 K)
Pressure = 100.0 KPa (100.0/101 = 0.98 atm)
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/molar mass
Number of moles = 6.40 g/ 16 g/mol
Number of moles = 0.4 mol
Now we will compare the moles of CO₂ with CH₄.
CH₄ : CO₂
1 : 1
0.4 : 0.4
Volume of CO₂:
Formula:
PV = nRT
0.98 atm ×V = 0.4 mol ×0.0821 atm.L/mol.K × 308 K
0.98 atm ×V = 10.11 atm.L
V = 10.11 atm.L /0.98 atm
V = 10.3 L
Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

z = 
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C =
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
You can boil or evaporate the water and the salt will be left behind as a solid. If you want to collect the water, you can use distillation. This works because salt has a much higher boiling point than water. One way to separate salt and water at home is to boil the salt water in a pot with a lid. So, I would say maybe oil.