Answer:
0.375 L
Explanation:
We know that at neutralization, the number of mol of acid must equal the number of equivalents of base.
This is a reaction 1:1 acid to base:
HClO₄ + NaOH ⇒ NaClO₄ + H₂O
We re given the moles of the base indirectly since we know the volume and molarity. From there we can calculate the volume of HClO₄.
Moles NaOH = 0.115 L x 0.244 M = 0.115 L x 0.244 mol/L =0.028 mol
Thus we require 0.028 mol of HClO₄ in the pechloric acid solution:
Molarity = # moles / V ⇒ V = # moles / M
V = 0.028 mol / 0.0748 mol/L = 0.375 L
Note that this problem can be solved in just one step since
M(HClO₄) x V(HClO₄) = M(NaOH) x V(NaOH) ⇒
V(HClO₄) = M(NaOH) x V(NaOH) / M(HClO₄)
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:
to launch aerials and also causes the explosions necessary for special effects like noise or colored light.
Explanation:
