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:
Reaction 1 is balanced but 2 is not balanced , the balance equation are :
1.
2.
Explanation:
Balanced Equations : These are the equation which follows the law of conservation of mass .
The total number of atoms present in reactant is equal to total number of atoms present in product.
1.
This is acid - base type reaction where
act as Acid
act as weak base
Reactant : ,
Number of atoms of :
C = 2 () + 1 (
)
= 2 + 1
= 3
H = 4() + 1 (
)
= 4 + 1
5
O = 2() + 3 (
)
= 5
Na = 1 ()
= 1
Product : ,
,
Number of atoms :
C = 1() + 2(
)
= 1 + 2
= 3
H = 2() + 3(
)
= 2 + 3
= 5
O = 1() + 2(
)
+2(
= 1 + 2 + 2
= 5
Na = 1(
= 1
Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction
2.
This is double displacement reaction .
Check the balancing in both reactant and products should be :
Na = 2
H = 2
Ca = 1
C = 2
O = 6
Cl = 2
Explanation:
According to the given data, we will calculate the following.
Half life of lipase = 8 min x 60 s/min
= 480 s
Rate constant for first order reaction is as follows.
=
Initial fat concentration = 45
= 45 mmol/L
Rate of hydrolysis = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration (S) =
= 45 (1 - 0.80)
= 9
or, = 9 mmol/L
It is given that = 5mmol/L
Therefore, time taken will be calculated as follows.
t =
Now, putting the given values into the above formula as follows.
t =
=
=
= 27.38 min
Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.