Answer: In the given reaction increasing the amount of solution is likely to increase the rate of the reaction.
Explanation:
It is known that more is the number of reactant molecules taking part in a chemical reaction more will be the number of collisions occur. As a result, more will be the rate of chemical reaction.
For example, When hydrogen peroxide dissociates into water and oxygen on addition of manganese oxide then increasing the number of reactants (hydrogen peroxide and manganese oxide) will also lead in the increase in rate of reaction.
Thus, we can conclude that in the given reaction increasing the amount of solution is likely to increase the rate of the reaction.
Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.
Answer:
The final volume is 39.5 L = 0.0395 m³
Explanation:
Step 1: Data given
Initial temperature = 200 °C = 473 K
Volume = 0.0250 m³ = 25 L
Pressure = 1.50 *10^6 Pa
The pressure reduce to 0.950 *10^6 Pa
The temperature stays constant at 200 °C
Step 2: Calculate the volume
P1*V1 = P2*V2
⇒with P1 = the initial pressure = 1.50 * 10^6 Pa
⇒with V1 = the initial volume = 25 L
⇒with P2 = the final pressure = 0.950 * 10^6 Pa
⇒with V2 = the final volume = TO BE DETERMINED
1.50 *10^6 Pa * 25 L = 0.950 *10^6 Pa * V2
V2 = (1.50*10^6 Pa * 25 L) / 0.950 *10^6 Pa)
V2 = 39.5 L = 0.0395 m³
The final volume is 39.5 L = 0.0395 m³
