Answer:
Molarity NaOH = 0.85M (2 sig figs)
Explanation:
48.0ml(0.220M H₂SO₄) + 25ml(Xmolar NaOH)H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
2(molarity x volume) H₂SO₄ = (molarity x volume) NaOH
2(0.220M x 48.0ml) = 25.0ml x Molarity NaOH
Molarity NaOH = 2(0.220M x 48.0ml)/25.0ml = 0.8448M ≅ 0.85M (2 sig figs)
Explanation:
hope it helps you with the question
Weak acids only partially ionize. so the retain part of the original molecule. C
Answer:
a. 52.8
Explanation:
To find the number of moles of HCl we use the relation M₁V₁=M₂V₂
where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.
M₁=7.91 M
M₂=2.13 M
V₁=?
V₂=196.1 mL
Replacing these values in the relationship.
M₁V₁=M₂V₂
7.91 M× V₁=2.13 M×196.1 mL
V₁=(2.13 M×196.1 mL)/7.91 M
=52.8 mL
Answer: The potential of the following electrochemical cell is 1.08 V.
Explanation:
=-0.74V[/tex]
=0.34V[/tex]
The element with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials, when concentration is 1M.
![E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BCu%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Thus the potential of the following electrochemical cell is 1.08 V.
