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>
3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and the last O atoms
Reaction
Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O
aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O
Zn : left = a, right =1 ⇒a=1
H : left = b, right = 2d⇒ b=2d (eq 1)
N : left = b, right = 2+c⇒b=2+c (eq 2)
O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)
3(2d)=6+c+d
6d=6+c+d
5d=6+c (eq 4)
3(2+c)=6+c+d
6+3c=6+c+d
2c=d (eq 5)
5(2c)=6+c
10c=6+c
9c=6
c = 2/3
d = 2 x 2/3
d = 4/3
b = 2 x 4/3
b = 8/3
The equation
aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to
Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3
3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O
Answer:
A- Physical, B- Chemical, C- chemical, D- Physical
Explanation:
A is physical because you can see it changing its form is changing.
B is Chemical because a new substance is formed creating the orange color of rust.
C is a Chemical reaction because it is being broken down so the banana itself is changing not just how we see it.
D is physical because we are just changing the shape/ size of the item, not anything to do with its substances.
Answer:
<u>a nonspontanenous reaction is forced to occur</u>
<em>Hope</em><em> this</em><em> helps</em><em> </em><em>:</em><em>)</em>
Explanation:
(A) As we know that carbonic acid (
) and Sodium bicarbonate (
) forms an acidic buffer.
Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.
pH = ![pK_{a} + log(\frac{[Salt]}{[Acid]})](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D%29)
........... (1)
So mathematically, if [Salt] = [Acid] then ![\frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
= 1
.
And, ![log (\frac{[Salt]}{[Acid]})](https://tex.z-dn.net/?f=log%20%28%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D%29)
= 0
Therefore, equation (1) gives us the following.
pH = 
(when acid and salt are equal in concentration)
Hence, of (carbonic acid) is 6.35.
And, with this we have following results.
In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].
Hence, for the cases pH = = 6.35.
(B) ![[NaHCO_{3}]](https://tex.z-dn.net/?f=%5BNaHCO_%7B3%7D%5D)
= 0.045 M and, ![[H_{2}CO_{3}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DCO_%7B3%7D%5D)
= 0.45 M
Hence, pH = ![6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])](https://tex.z-dn.net/?f=6.35%20%2B%20log%28%5BNaHCO_%7B3%7D%5D%5B%5BH_%7B2%7DCO_%7B3%7D%5D%29)
= 
= 6.35 + (-1)
= 5.35
Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.
(C) = 0.45 M
= 0.045 M
So, pH = ![6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})](https://tex.z-dn.net/?f=6.35%20%2B%20log%28%5Cfrac%7B%5BNaHCO_%7B3%7D%5D%7D%7B%5BH_%7B2%7DCO_%7B3%7D%5D%7D%29)
= 
= 6.35 + (+1)
= 7.35
This means that pH on Basic side makes it no more acidic buffer.
