Answer:
d
Explanation:
Firstly, we need to see the theoretical mole ratio between nitrogen and ammonia from the balanced chemical equation. This is 1 to 2. One mole of nitrogen yielded two moles of ammonia.
At STP, one mole of a gas occupies a volume of 22.4, hence we need to know the volume occupied by a volume of 44.8L of ammonia. This is equal to 44.8/2 = 2 moles
Now we have seen the actual number of moles of ammonia yielded. Since this is the same as the theoretical, it means that only one mole of nitrogen was also used up.
Since it is one mole, the volume at STP is thus 22.4L
Answer:
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Explanation:us8s f7 ds8a das8 jf7sd 87sd 45s67f8
Answer:
9
Explanation:
let f(x)= x-7 be equation i
let g(x)= x^2 be equation ii
Substitute x-7 as x in equation 2
implies gf(x) = (x-7)^2=(x-7)(x-7)
substitute x=4 implies
gf(x) = (4-7)(4-7)=(-3)(-3)=9
I can give the answer for the first one: Because the football is being thrown into t
Answer: "Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical equation and tables displaying information about enthalpy of formation/combustion. Write out the solution according to Hess's law." That's all I'm given. The key says that the answer is -125.4 kJ. C4H10 + 6.5 O2 = 4CO2 + 5 H20 The heat of combustion for one mole of butane is -2877 K. Heats of formation from table C02 = -393.5 Kj/mole, H20 (g) -241.8 Kj/mole. Heat of combustion of butane = sum heats of formation products minus heats of formation reactants -2877 Kj = 4(-393.5Kj) + 5(-241.8Kj) minus heat or enthalpy of formation butane.
The enthalpy formation oxygen an element is assigned a value of zero.
-2877 Kj = -1582 Kj + -1209 Kj minus X
-2877 = - 2791 Kj minus X
X = -86 Kj. The heats of formation for various compounds vary a little from table to table, so this does not quite agree with your answer.
This is just an example
