13
Al
Aluminum
Atomic mass 26.982
Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Solution : Given,
Mass of Cu = 300 g
Molar mass of Cu = 63.546 g/mole
Molar mass of 
= 183.511 g/mole
- First we have to calculate the moles of Cu.
The moles of Cu = 4.7209 moles
From the given chemical formula, we conclude that the each mole of compound contain one mole of Cu.
So, The moles of Cu = Moles of = 4.4209 moles
- Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of = 4.4209 moles × 183.511 g/mole = 866.337 g
Mass of = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)
Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.
Answer:
36.8 L
Explanation:
We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 80 °C
T(K) = 80 + 273
T(K) = 353 K
Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:
Number of mole (n) = 1.27 moles
Temperature (T) = 353 K
Pressure (P) = 1 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
1 × V = 1.27 × 0.0821 × 353
V = 36.8 L
Thus, the volume occupied by the helium gas is 36.8 L
Sorry the answer might be a little late but it would be <span>Simple squamous</span>
