Answer:
= 67.79 g
Explanation:
The equation for the reaction is;
4Cr(s)+3O2(g)→2Cr2O3(s)
The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;
moles O2 = 21.4 g / 32 g/mol
=0.669 moles
Using mole ratio, we get the moles of Cr2O3;
moles Cr2O3 = 0.669 x 2/3
=0.446 moles
but molar mass of Cr2O3 is 151.99 g/mol
Hence,
The mass Cr2O3 = 0.446 mol x 151.99 g/mol
<u> = 67.79 g
</u>
Given that the pressure, temperature and area of effusion is constant, the rate of effusion is inversely proportional to the square root of the molecular mass of the gas.
Mr Oxygen = 32
Mr Argon = 40
Effusion Oxygen = 1/√32
Effusion Argon = 1/√40
Effusion Oxygen / Effusion Argon = √(40) / √(32)
=√(40/32) = √(5/4) = 1.19
Thus, Oxygen will effuse 1.19 times faster than Argon. The second option is correct.
Answer:
0.123 moles of ammonia, can be produced
Explanation:
First of all, we need to determine the reaction:
Ammonia is produced by the reaction of hydrogen and nitrogen.
3H₂(g) + N₂(g) → 2NH₃(g)
Ratio is 3:2. Let's solve the question with a rule of three:
If 3 moles of hydrogen can produce 2 moles of ammonia
Then, 0.37 moles will produce (0.37 . 2) /3 = 0.123 moles
NH4I (aq) + KOH (aq) in chemical equation gives
NH4I (aq) + KOH (aq) = KI (aq) + H2O(l) + NH3 (l)
Ki is in aqueous state H2o is in liquid state while NH3 is in liquid state
from the equation above 1 mole of NH4I (aq) react with 1 mole of KOH(aq) to form 1mole of KI(aq) , 1mole of H2O(l) and 1 Mole of NH3(l)
I would say C. petroleum
A is wrong
B is impractical for "mainly used"
D thats too expensive
