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
Answer:
The correct answer is option B
Explanation:
Given values,
Molarity of 
Volume of solution, 
Molecular weight of 
Substituting this values in Molarity formula, we get
Use PV=nRT to find V assuming n is one and R= 8.31 then use the answer to find P2
Answer:
A = 0.75 ×10² KJ.
B = 3.9 ×10³ dg
C = 0.22 × 10² μl.
Explanation:
A = 7.5 ×10⁴ j to kilo joules
7.5 ×10⁴ / 1000 = 0.75 ×10² KJ.
Joule is the smaller unit while kilo joule is the larger unit. One kilo joule equals to the thousand joule that's why we will divide the given value by 1000 in order to convert into KJ.
B = 3.9 ×10⁵ mg to decigrams.
3.9 ×10⁵ / 100 = 3.9 ×10³ dg
Decigram is larger unit while milligram is smaller unit. One decigram is equal to the 100 milligram. In order to convert the given value into decigram we have to divide the value by 100.
C = 2.21 ×10⁻⁴ dL to micorliters
2.21 ×10⁻⁴ ×10⁵ = 0.22 × 10² μl.
Deciliter is bigger unit then micro liter . One deciliter equals to the 100000 micro liters. In order to convert the dL into micro liter we have to multiply the given value with 100000.
The balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of N₂ to H₂ is 1:3
we need to find the moles of N₂, volume of N₂ has been given
molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP.
if 22.4 L is occupied by 1 mol
then 3.5 L of gas is occupied by - 3.5 L / 22.4 L/mol = 0.16 mol
number of moles of N₂ present - 0.16 mol
1 mol of N₂ requires 3 mol of H₂
therefore 0.16 mol of N₂ requires - 3 x 0.16 = 0.48 mol of H₂
mass of H₂ required - 0.48 mol x 2 g/mol = 0.96 g
0.96 g of H₂ is required
