Sophomore Chemistry Questions 15, Sophomore Chemistry Questions

10.If in 25, pH = 2 H2SO4, then C (H +) = PH = 11 NaOH solution then POH = 14 - 11 = 3, then C (OH-) = acid and base neutralization, mixing solution pH = 7

then n(h+)=n(oh-).

c(h+) v(h+) = c(h+)*v(oh-) So, v(h+): v(oh-) = 1:10

Then the volume ratio of the two solutions is H2SO4:NaOH 1:10 is a strong electrolyte, which is completely ionized in the solution.

AGCL – AG+ CL- (should be a double arrow) The order of concentration of AG+ from large to small, which is a question about ionization equilibrium.

10 ml in water.

CaCl2 solution, n(Cl-)=

KCl solution, n(Cl-)=

NaCl solution, n(Cl-)=

The larger the n(Cl-), the more ionization is inhibited, the more difficult it is to ionize AgCl, and the less Ag+.

So a ①

It should be detailed enough! The unit is so difficult to fight, huhu is tired].

10,n(oh-)=n(h+)

So v(H2SO4)*(10 -2)*2=v(Naoh)*(10 -3).

Therefore, V(H2SO4):V(NaOH)=1:1011, the greater the Cl- concentration, the more it inhibits the dissolution of AGCL.

The CL- concentration size is 3>2>4>1

Therefore, the concentration of AG+ from large to small is 1>4>2>3 choose B

The concentration of H2SO4 hydrogen ions with pH=2 is .

The hydroxide concentration of NaOH solution with pH=11 is.

pH = 7, completely neutralized, then the amount of the substance of the hydrogen ion and the amount of the substance of the hydroxide group are equal.

So it's 1:10

The answer should be d, it should be no diagram, you can only think about it.

The copper in A is obviously the negative electrode, and the oxidation reaction occurs, and the mass of the silver electrode in B gradually increases, and C seems to be correct, but the electrolyte solution is the directional movement of anions and cations, and there is no electron transfer, so the statement is wrong.

So choose D

The electrons of the positive electrode, the negative electrode loses electrons.

Cu-2E-= Cu2+ negative Aag++E-=Ag Positive B Y Reduction of C(AG+) in solution.

x(cu) electrode mass decreases electron transfer in electrolyte solution c electrochemical reaction - galvanic cell so the pointer of the galvanometer in the external circuit is deflected d cd

d correct. a: The X pole loses electrons, so it is a negative pole; b: ag e-==ag, so the concentration of silver ions in the solution decreases; C: The charge transfer in the electrolyte solution is done by anions and cations, so there is no electron transfer.

DA is wrong, it should be the negative electrode, B is wrong, the mass of the silver electrode increases, C(AG+) decreases, C is wrong, and there is electron transfer in the external circuit.

The correct answer is D, welcome to ask.

The silver electrode is the positive electrode of the battery, and the electrode reaction that occurs is that the concentration of Ag+ in the ag++e-=ag solution gradually decreases.

x is the negative pole. The answer is cd

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