Q. No. 1:
a) Endothermic
b) Exothermic
c) Endothermic
d) Exothermic
Q. No. 2:
a) Exothermic
b) Because heat is on product side (evolved to surrounding).
c)
1400 Kj = was produced by 1 mole of C₂H₄
Then,
2100 Kj = will be produced by X moles of C₂H₄
Solving for X,
X = (2100 Kj × 1 mol) ÷ 1400 Kj
X = 1.5 moles of C₂H₄
d)
3 moles of O₂ produced = 1400 Kj of Heat
Then,
1.5 moles of O₂ will produce = X Kj of Heat
Solving for X,
X = (1.5 mol × 1400 Kj) ÷3 mol
X = 700 Kj of Heat
Answer:
1dm = 10cm
1dm3 = 10cm x 10cm x 10cm = 1000cm3
1L = 1000cm3
so 1litre = 1dm3
4 moles in 2 dm3
= 2 moles in 1dm3 = 2moles in 1L = 2M
1L = 1000cm
0.3 moles in 200cm3 =
200cm3 = 0.2L
so 0.3 moles in 0.2 L = 1.5 moles in 1 L = 1.5M
(I have tried to simplify and show why 1dm3 = 1L)
I hope this helps you understand!
Answer:
I dont know the answer for that question it's hard question isn't it
Answer:
Oxygen is the limiting reactant.
Explanation:
Hello,
In this case, given the reaction:
Hence, given the masses of both ethanol and oxygen, we are able to compute the available moles ethanol by:
Next, we compute the moles of ethanol that react with the 0.640 grams of oxygen considering their 1:3 molar ratio in the chemical reaction:
In such a way, since there are 0.01 available moles of ethanol but just 0.0067 moles are reacting, we evidence ethanol is in excess, therefore the oxygen is the limiting reactant.
Best regards.