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: The mass of deposited cadmium is 23.9g
Explanation:
According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode is directly proportional to the quantity of electricity passed through the electrolyte.
Faraday has found that the amount of electricity needed to liberate one gm eq. of substance from an electrolyte is 96500C.
Given that
time (t) =1.55min *60 =93seconds
Current (I) =221A
Q= It = 221 * 93 =20553C
Molar mass of CdS04 =112.41
96500C will liberate 112.41g
20553C will liberate Xg
Xg= (20553*112.41)/96500
=23.9g
Therefore , the amount of Cd deposited at the electrode is 23.9g
The net ionic equation is
Ag⁺(aq) +Cl⁻(aq) → AgCl(s)
<u><em>Explanation</em></u>
AgNO₃ (aq) + KCl (aq)→ AgCl(s) +KNO₃(aq)
from above molecular equation break all soluble electrolyte into ions
Ag⁺(aq) +NO₃⁻ (aq) + K⁺(aq) +Cl⁻(aq) → AgCl (s) + K⁺(aq) + No₃⁻(aq)
cancel the spectator ions in both side of equation =K⁺ and NO₃⁻ ions
The net ionic equation is therefore
= Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
This is an incomplete question, here is a complete question.
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kJ. You may want to reference (Pages 381 - 385) Section 9.6 while completing this problem. If the change in enthalpy is -5074.2 kJ, how much work is done during the combustion? Express the work in kilojoules to three significant figures.
Answer : The work done during the combustion is, 9.9 kJ
Explanation :
Formula used :
where,
w = work done = ?
= change in enthalpy = -5074.2 kJ
= change in internal energy = -5084.1 kJ
R = gas constant = 8.314 J/mol.K
Now put all the given values in the above formula, we get:
Thus, the work done during the combustion is, 9.9 kJ
Answer: Tropical rain forests are warm and wet, and the temperature does not vary much. Because of this, a huge variety of plants can live in tropical rain forests. Because there are so many kinds of plants, there are habitats for many different kinds of animals as well.
Explanation: