Junior high school chemistry questions, urgent help, junior high school chemistry questions, help

Assuming that the metal mixture can be expressed in m (and that 2-valence ions are formed in solution, note that this is only a hypothesis), the equation for the reaction can be written as:

m＋h2so4＝mso4＋h2

According to the title, the mass of MSO4 obtained in the end is b grams, and M has a gram, and the increased mass is the mass of the sulfate participating in the reaction, so that the mass of the sulfate participating in the reaction is (b a) grams, and the amount of the substance is (b a) 96mol, which is the amount of sulfuric acid, that is, the amount of hydrogen generated.

b a) 96mol, hydrogen mass (b a) 48 g.

Since the salt produced is sulfate, its mass is the mass of the metal plus the mass of the sulfate, i.e.:

b = x + a (x is the mass of sulfate).then x=b-a

According to the relationship between the mass of H2 generated and the sulfate:

h2so4---h2---so4

You can calculate the mass of H2. Namely:

h2---so4

y---b-a

y = 2 (b-a) 96 = (b-a) 48 (g).

The third one is clear and simple to understand. The second one seems to be out of the ordinary. I didn't talk about "the quantity of matter" in junior high school!

k：i：o=39：127：48

Mass fraction of i=127 (39+127+48)mass of NaOH 125*10%=

The second part of the measurement of water is that is.

Part 3 Add NaOH and stir to dissolve.

Is hydrochloric acid by lz hydrochloric acid mass fraction? I'll just do that.

hcl + naoh=nacl+h2o

50* xx=40*50*

The mass fraction of NaOH in the remaining solution is 4 (125-5)*100%=

The ratio of iodine to oxygen to mass is: 127:16*3=127:

The mass fraction of 48 iodine element is 127 (39+127+16*3)=25The mass of NaOH is: 125*10%= 2

Weighing. 3。Dissolve.

naoh + hcl ==nacl + h2ox=4g 50*

Mass fraction of NaOH in the remaining solution: 4 (125-5)=

25 (1)12,5, 2 Weigh NaOH 12,5g with a balance,3 Pour 12,5g of NaOH solid into water to dissolve.

2）naoh ＋ hcl = nacl ＋ h2om（naoh） 50g×

m（naoh）=4g

naoh）=4g/120g ×100％=

First of all, let's ask, what is naco3?

The solution mitigates grams, i.e. the CO2 released by the reaction, altogether. Therefore, the original solid Na2CO3 has, i.e., grams.

The final solution contains a total gram of NaCl, so the mass fraction is.

Solubility s=

na2co3 + 2hcl= 2nacl + h2o + co2

x y 11 + ie.

x= y= so the mass fraction of the solute is (

The content of Na2CO3 in the original mixture is as good as the quality.

Solubility of the solution at 30 degrees Celsius.

s=The key to this problem is to make good use of the reduced mass of carbon dioxide to find the quality of sodium carbonate in the original mixture and the mass of sodium chloride generated by the reaction, and then you can find the sodium chloride in the original mixture, so as to find the total sodium chloride mass after the reaction, and the next problem will come out naturally!

This question involves the law of conservation of mass, you have to take a good look at the original words in the textbook, this law is very useful! You love learning so much! Come on, everyone will try their best to help you if you have any questions!

There is one that can be dissolved.

There is a dissolvable substance.

nacl+agno3=nano3+agclmgcl2+2agno3=mg(no3)2+2agclbacl2+2agno3=2agcl+ba(no3)2alcl3+3agno3=al(no3)3+3agcl2kcl+agco3=2agcl+k2co3na2so4+bacl2=2nacl+baso4cuso4+baco3

K2SO4 + BA(NO3)2 = 2KNo3 + BASO4Na2CO3 + BA(NO3)2 = 2nano3 + BAC3K2CO3 + CA(NO3)2=CaCO3 + 2KNO3 A soluble.

There is a dissolvable substance.

6. The properties of ammonium salt are unstable, and ammonia can be decomposed when exposed to alkali.

Ammonium sulfate, ammonium nitrate.

Ammonium chloride ammonium phosphate.

You are empty and dust-resistant, and you are very happy to answer for you.

I don't know how to ask thank you.