One chemistry question with extra points Two chemistry questions, extra points

It's relatively simple, start with aluminum ions, because aluminum ions are moles, so the corresponding Al2(SO4)3 is moles, so the sulfate ions in aluminum sulfate have moles.

And because there are molar sulfate ions in total, the sulfate in sodium sulfate accounts for moles.

So the sodium ion is moles.

The negative ions in the solution are only sulfate ions, 2, the positive ions are only aluminum ions and sodium ions, and the aluminum ions are 3, and the charge in the solution is conserved, so the sodium ions should be 1, so the sodium ions are moles.

The anions in the solution are only sulfate ions, and the cations are aluminum ions and sodium ions, and according to the principle of ion conservation, the number of anions and cations should be equal. Al2(SO4)3 in the aluminum ion (+3) is, then the sulfate ion (+2) is, then the sulfate ion in Na2SO4 is there, then the sodium ion (+1) has 1mol

Sulfate 2-valent so2*, aluminum ion trivalent so, so sodium ion is 2*

For, let Na2SO4 X parts, Al2SO4 Y parts, isn't it OK, those two do not react, X+3Y=

2y= na is 2x

1. Let the volume of H2 x and the volume of F2 be y

1) H2 reacts exactly with F2.

x=y=a/2

b=02) H2 excess, then b is excess H2

x+y=ab=x-y

The solution is that x=(a+b) 2, y=(a-b) 2--- i.e., option b.

3) If there is an excess of F2, then B is the O2 generated after the reaction of the excess F2

x+y=ay-x)/2=b

Solution, x=(a-2b) 2, y=(a+2b) 2

This is not the case in the options.

c2h4o2 + 2o2 = 2co2 + 2h2o

ch2o + o2 = co2 + h2o

It can be found that the reaction of 1mol of two substances with O2 is O2 deficiency.

The resulting CO2, H2O, and Hui can react with Na2O2.

2co2 + 2na2o2 = 2na2co3 + o2

2h2o + 2na2o2 = 4naoh + o2

It can be found that there is an excess of Na2O2.

Therefore, organic matter is burned, and the CO2 and H2O produced react with the excess Na2O2 to form O2

The amount of gas in the container is equal to the amount of merging substances of the gas before the reaction--- i.e., the O2 generated later is equal to the O2 consumed earlier.

That is, the composition of the organic matter is in accordance with (co)mhn, so both organic matter are in accordance with it.

However, CH2O is also a gas at room temperature and is also counted in the gas before the reaction, so it is not compliant.

This question comes from the classic, M g organic matter (Co) XHY, the gas after full combustion reacts with excess Na2O2, and the solid mass increases M G, which is examined from the perspective of O2.

However, the problem of answering the question in this draft is that when organic matter is burned, the amount of O2 is insufficient, which will lead to the formation of CO and other gases in organic matter, so it cannot be analyzed in this way.

1) The two cases are discussed: when the remaining gas is O2, the volume of F2 in the original gas mixture is (A+2B) 2; When the remaining gas is H2, F2 in the original gas mixture is option B.

2) I haven't thought too much about Lu Youchu, but HCO is definitely not right, because it is gaseous at room temperature and pressure. CH3COOH can be sold early to react with Na2O2.

,)=1/2li2o(s) △h(

2.Negative electrode. Alkali metals react with water.

508g, <5> power c

1）4li+o2=2li2o

2) The cathode li undergoes a displacement reaction with water 254

If the electrolyte solution is an alkaline solution, the negative electrode reaction formula is: 2H2 + 4OH -4E ==4H2O Positive Electrode: O2 + 2H2O + 4E ==4OH

That is, A is the positive electrode and B is the negative electrode. The end of Cu is the negative pole and Zn is the positive pole. When zn is the negative electrode, it loses electrons and its mass decreases. In fact, the galvanic cell on the right did not happen, so the concentration of CuSO4 remained unchanged, and D was chosen

a.The beaker on the left is the galvanic cell, and on the right is the electrolytic cell. So cu is the anode, and it cannot be said that it is positive.

b.Cu2+ in solution is precipitated on the Zn sheet, and Zn itself is not involved in the reaction CThe solution environment is alkaline, so it should be written as H2 2E- +2OH- =2H2O

d。That's right.

Reasonable is d.

The reaction of the A pole in the hydrogen-oxygen fuel cell is O2+4E+2H2O=4OH, because the electrolyte in the galvanic battery is KOH solution, there will be no 2H+ generation, and the B pole should be H2-2E+2OH=2H2O, so A and B are the positive and negative electrodes of the galvanic cell, and correspondingly, the Cu and Zn electrodes are the anode and cathode of the electrolytic cell. The electrolyte in the electrolytic cell is CuSO4 solution, the Cu anode loses electrons: Cu-2E=Cu2+, and the Zn cathode gains electrons:

Cu2+ +2e=Cu, so the mass on the zinc sheet increases due to the enrichment of Cu.

(1) 6 (solute) (6 + 114) (total solution mass) = 5% (2) 6 (solute) (6 + 114 + 30) (total solution mass) = 4% (3) (6 + 5) (solute) (6 + 114 + 5) (total solution mass) = (4) Let the water be added as y, then there is 6 (solute) (6 + 114 + y) (total solution mass) = solution y = 120

5) Let the evaporated water mass be x, then there is 6 (solute) (6 + 114-x) (total solution mass) = 10%, and the solution is x = 60

(1) 6 divided by (114 + 4) = 5%.

2) 6 divided by (114 + 6 + 30) = 4%.

3) (6 + 5) divided by (114 + 6 + 5) =

4) 6 divided, 240 - (114 + 6) = 120 (5) 6 divided by 10% = 60, 114 - 60