The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
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.
Yes it is ========== covalent bond
Answer:
31.5mL
Explanation:
The following were obtained from the question:
C1 (concentration of stock solution) = 2M
V1 (volume of stock solution) =.?
C2 (concentration of diluted solution) = 0.630M
V2 (volume of diluted solution) = 100mL
Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
2 x V1 = 0.630 x 100
Divide both side by 2
V1 = (0.630 x 100) /2
V1 = 31.5mL
Therefore, 31.5mL of 2M solution of FeCl2 required
