Answer:
Final temperature =78°C
Explanation:
The amount of heat lost is calculated using the formula for calculating the enthalpy change: mCΔT C, the specific heat capacity for for water is 4.186J/gK. The mass of water is 30 grams.
1500J= 30g×4.186J/gK×ΔT
ΔT=1500J/(30×4.186J/gK)
=11.94K
final temperature=(90-12)°C
=78°C
Answer:
The right solution is "
".
Explanation:
As we know,
1 mole electron = 
Total energy = 
= 
For single electron,
The amount of energy will be:
= 
= 
Answer:
92 ATP
Explanation:
Fatty acid oxidation results in the formation of large number of ATP molecules. Three important process of fatty acid are activation of the fatty acid, beta oxidation and entry of acetyl CoA in Krebs cycle.
14 carbon fatty acid is Miristic acid. The complete oxidation of Miristic acid results in the formation of 7 acetyl CoA + 6NADH and 
1 Acetyl CoA gives 10 molecules of ATP then 7 acetyl CoA gives 70 molecules of ATP.
1NADH = 2.5 ATP, 6NADH = 15 ATP.
= 1.5 ATP,
= 9 ATP.
2 ATP has been consumed in the activation of fatty acid.
Total ATP = 70+15+9-2
=92 ATP.
Thus, the total ATP generated from the oxidation of 14 carbon fatty acid is 92.
Answer : The complete chemical equation is,

Explanation :
As we know that, in a chemical equation the reacting species present on left side and the product formed present on right side and a right arrow inserted between the reactants and product that show a chemical reaction taking place.
In the chemical reaction, the phases of the substances are also included and subscripts and superscripts are also used for the numbers.
For the given chemical reaction, the balanced chemical equation including the phases, is given by:

Reactions of Ethyl-3-pentenoate with all given reagents are given below.
Reaction with H₂ / Pd:
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.
Reaction with NaBH₄: Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.
Reaction with LiAlH₄: Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.
Result: The correct answer is
Option-A (Highlighted RED below).