Answer : The correct option is, (D) 3600 kJ
Explanation :
Mass of octane = 75 g
Molar mass of octane = 114.23 g/mole
Enthalpy of combustion = -5500 kJ/mol
First we have to calculate the moles of octane.

Now we have to calculate the heat released in the reaction.
As, 1 mole of octane released heat = -5500 kJ
So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)
= -3608 kJ
≈ -3600 kJ
Therefore, the heat released in the reaction is 3600 kJ
<span>(2.09 mL) x (1.592 g/mL) / (227.0871 g C3H5O9N3/mol) = 0.014652 mole C3H5O9N
4 moles C3H5O9N produce 12 + 6 + 1 + 10 = 29 moles of gases, so:
(0.014652 mole C3H5O9N) x (29/4) = 0.106 mole of gases
(b)
(0.106 mol) x (46 L/mol) = 4.88 L gases
(c)
(0.014652 mole C3H5O9N) x (6/4) x (28.0134 g/mol) = 0.616 g N2</span>
I believe you just look at your periodic table for this value. I don't think there is any math involved.
Therefore one mole of Mg = 24.305g.
Answer:
Explanation:
We want the energy required for the transition:
CO 2
(
s
)
+
Δ
→
C
O
2
(
g
)
Explanation:
We assume that the temperature of the gas and the solid are EQUAL.
And thus we simply have to work out the product:
2
×
10^
3
⋅
g
×
196.3
⋅
J
⋅
g
−
1 to get an answer in Joules as required.
What would be the energy change for the reverse transition:
C
O
2
(
g
)
+
→
C
O
2
(
s
)
?
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234





So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left

Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.