<u>Answer:</u> Copper is getting oxidized and is a reducing agent. Silver is getting reduced and is oxidizing agent.
<u>Explanation:</u>
Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.
Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.
Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.
Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.
For the given chemical reaction:
The half reactions for the above reaction are:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u> 
From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.
Silver is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.
Answer:
Time take to deposit Ni is 259.02 sec.
Explanation:
Given:
Current 
A
Faraday constant 
Molar mass of Ni 
Mass of Ni 
g
First find the no. moles in Ni solution,
Moles of Ni 
mol
From the below reaction,
⇆ 
Above reaction shows "1 mol of 
requires 2 mol of electron to form 1 mol of "
So for finding charge flow in this reaction we write,
Charge flow 
C
For finding time of reaction,
Where 
charge flow
sec
Therefore, time take to deposit Ni is 259.02 sec.
Answer:
9 × 10⁻³ mol·L⁻¹s⁻¹
Explanation:
Data:
k = 1 × 10⁻³ L·mol⁻¹s⁻¹
[A] = 3 mol·L⁻¹
Calculation:
rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹
Answer:
700 calories
Explanation:
Using the formula below:
Q = m × c × ∆T
Where;
Q = amount of heat required (calories)
m = mass of substance (g)
c = specific heat of substance (cal/g°C)
∆T = change in temperature (°C)
According to this question, the following information was provided;
Q = ?
m = 20g
c = 1.0 cal/g °C
∆T = 40°C - 5°C = 35°C
Using the formula; Q = m × c × ∆T
Q = 20 × 1 × 35
Q = 700 calories
Hence, 700 cal of heat energy is needed to raise 20 g of H2O from 5°C to 40°C.
Answer:
Hey do you know if "horsleyjaydyn" is a girl you commented once she was a guy so
Explanation:
