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)
1. NaCl
2. CaO
3. KOH
4. MgS
5. CuCO3
6. Al2O3
7. Fe2O3
8. Na2CO3
9. AlHO3
10. (NH4)3N
11. Zn3N2
12. MgCO3
All numbers are subscripts.
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Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
