Chemistry questions about chemical formulas in junior high school

It should be (2*28): 44

It can be considered that if you want to contain the same mass of oxygen elements, (because they are the same elements), so you have to have the same number of oxygen atoms, then 2 CO molecules contain as many O atoms as 1 CO2 molecule, so the quantity ratio of CO and CO2 is 2:1, so the mass ratio is equal to the ratio of quantity multiplied by molecular weight, that is, 56:44, which is further simplified to 14:11

Let the CO2 containing xmol and ymol be the same, then.

16x=32y gives x:y=2:1

So the mass ratio is (12+16)x:(12+16*2)y=(28*2) (44*1)=14:11

The ratio of the number of c and o in co is 1:1, the mass ratio is 12x:16x=3x:4xco2, the ratio of c and o is 1:2, and the mass ratio is 12y:32y=3y:8y

The quality of the o among them is the same.

So: o has the same mass so 4x=8y

x:y=2:1

The mass of CO is 3x2+4x2=14

The mass of CO2 is 3x1+8x1=11

So the mass ratio should be 14:11

I guess that's the case.

After adding NaOH all Mg2+ is precipitated by OH- Part Ca2+ is bound to OH There is Ca2+ Na+ Oh-Cl- in the solution, so there is solute CaCl2 Ca(OH)2 NaCl NaOH After adding NaCO3, all Ca2+ in the solution is precipitated Na+ Oh- Cl- (CO3)2- So there is solute NaCl Naoh Na2CO3

After mixing, the mass of the solution decreases, and there must be precipitation or gas formation, so it is 2HCl+Na2CO3=2NaCl+H2O+CO2 Na2CO3+Ca(OH)2=CaCO3 +2NaOH

After mixing, the solute mass remained unchanged HCl+NaOH=H2O+NaCl 2HCl+Ca(OH)2=2H2O+CaCl2

Because the gas does not appear until some time later, there must be an OH- mixture in the solution, which may be composed of NaOH NaCl, Na2CO3 and Ca(OH)2 CaCl2 (in this case, there must be CaCO3 precipitation).

Changed one of the two naohs to ca(oh)2 I don't know if it's right.

1. Add excess sodium hydroxide, sodium hydroxide, calcium hydroxide.

Add excess sodium carbonate, sodium hydroxide, sodium carbonate.

Add dilute hydrochloric acid, hydrochloric acid (remove by heating at the end).

2 Which five solutions are there and why are there two sodium hydroxides?

1）2hcl+na2co3=2nacl+h2o+co22）naoh+hcl=nacl+h2o

3）naoh，na2co3

1 mgCl2+2NaOH = mg(OH)2+2NaCl The rest is CaCl2

CaCl2+Na2CO3=CaCO3+2NaCl What is left is excess sodium hydroxide.

After adding dilute hydrochloric acid, there is no residue only sodium chloride.

2 (1)2HCl+Na2CO3=2NaCl+CO22) Your question is wrong, is there four or not five, what is that one?

3) Sodium hydroxide and sodium carbonate, sulfuric acid reacts with sodium hydroxide first, and then reacts with sodium carbonate.

1. The first step is calcium hydroxide (slightly soluble) and sodium hydroxide (excess); In the second step, sodium carbonate (excess) and sodium hydroxide (non-reactive); In the third step, if the dilute hydrochloric acid is excessive, there is dilute hydrochloric acid, if the reaction is just complete, there is only sodium chloride, if it is insufficient, there is sodium carbonate or sodium carbonate and sodium hydroxide (in solution, it is generally regarded as other reactions after the neutralization reaction is completed).

Question 2: Repeat the five solutions and answer them after the change.

CaCl2+2NaOH = 2Na + Ca(OH)2 CaCo3 + Ca(OH)2= no reaction.

mgcl2+2naoh=nacl+mg(oh)2 caco3+mg(oh)2 =ca(oh)2+mgco3

Excess sodium hydroxide should not react with excess sodium carbonate.

2hcl+mgco3=h2co3+mgcl2 naoh+hcl=nacl+h2o na2co3+2hcl=2nacl+h2co3

1) 2HCl+Na2CO3=2NaCl+H2CO32)HCl+Naoh=NaCl+H2O, how to have one less solution.