Answer:
-973 KJ
Explanation:
The balanced reaction equation is;
N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)
Reduction potential of hydrazine = -1.16 V
Reduction potential of chlorine = 1.36 V
From;
E°cell= E°cathode - E°anode
E°cell= 1.36 - (-1.16)
E°cell= 2.52 V
∆G°=- nFE°cell
n= number of moles of electrons = 4
F= Faraday's constant = 96500 C
E°cell = 2.52 V
∆G°=- (4 × 96500 × 2.52)
∆G°= -972720 J
∆G°= -972.72 KJ
F₂ + 2 NaI → 2 NaF + I₂
<span>It is given that F₂ is light yellow / colorless in hydrocarbon solvent. The student combines Fluorine water with NaI in water. Then student adds pentane in the mixture of F₂ and NaI. After dissolution, solution was observed and a colorless pentane layer was seen. Alkanes are unreactive in nature. The C-H bond in alkane is difficult to break. whereas, F₂ is very reactive and reacts vigorously with alkanes in presence of light by free radical mechanism.It is given that the color of the solution is nearly colorless. F₂ when present in hydrocarbon solvent is light yellow/ colorless/ nearly colorless. Hence, F₂ is not reacting with hydrocarbon and there is no reaction taking place (No F</span>₂ is present<span>)</span>
V2 = 250 ml
Explanation:
Given:
P1 = 0.99 atm. V1 = 240 ml
P2 = 0.951 atm. V2 = ?
We can use Boyle's law to solve for V2
P1V1 = P2V2
V2 = (P1/P2)V1
= (0.99 atm/0.951 atm)(240 ml)
= 250. ml
Answer: 22g of chlorine would be needed to carry out this synthesis reaction
Explanation:
A synthesis reaction is one in which two or more than two elements combine together to forma single product.
The atoms present in the reactants are found on the product side. According to the law of conservation of mass, the number of atoms on both sides of the arrow must be same as the total mass must be conserved.
15 grams of sodium reacts with 22 grams of chlorine to yield 37 grams of sodium chloride. Thus 22g of chlorine would be needed to carry out this synthesis reaction.
6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.
0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.
