Chemistry questions in junior high school, all the bosses can help!

It should be put in slowly from top to bottom, because if it is quickly put to the bottom, it will be locally exothermic, difficult to dissipate, and easy to cause.

You can put it in quickly... But no obvious effect was observed... Your laboratory is going to observe the phenomenon, so it's better to put it in slowly!!

I don't know why, but I think the carbon monoxide produced by the contact between red-hot charcoal and oxygen is denser than the oxygen bottom, and floats upward.

Because the bottle is filled with pure oxygen, if it is slowed down, the charcoal will burn violently in the pure oxygen, and it is likely that the spoon and the bottle will be burned, and the charcoal in the oxygen can burn to more than 2000 degrees.

Slowly put the charcoal down from the top of the bottle. The reason is that the oxygen is evenly distributed in the bottle, and the charcoal should be allowed to react with the oxygen at the mouth of the bottle first, and then slowly move to the bottom of the bottle, so that the charcoal can fully react with the oxygen, and the high-temperature hot gas after the reaction will flow out of the bottle without affecting the oxygen in the lower part; If it is directly inserted into the lower part of the bottle, the hot gas produced by the reaction of the charcoal with the oxygen will push the oxygen in the bottle out of the bottle. So that the reaction stops quickly.

I don't know what you're talking about, but I know the reason, which is that if the charcoal burns incompletely, it will produce carbon monoxide, which is gas, and carbon monoxide is lighter than oxygen, so if your nose is just above the mouth of the test tube, it will be easy to be poisoned.

From top to bottom, the reason is that the oxygen is evenly distributed in the bottle, and it should first react with the oxygen at the mouth of the bottle, and then slowly move to the bottom of the bottle, if it reaches into the lower part of the bottle, the oxygen at the mouth of the bottle floats away and causes waste.

The title is not clear, "should be slowly put in from --- to ---" means charcoal and oxygen?

Let's go from the top down. Can you put it from the bottom up?

d k2cr2o7

Potassium +1 valence, oxygen-2 valence;

cr2(so4)3

Sulfate-2 valence.

①.Mass fraction of iron in Fe2O3.

56*2/(56*2+16*3)*100%=70%②。800t of hematite containing 60% Fe2O3. Theoretically, how many tons of pure iron can be smelted.

According to the equation.

Fe2O3+3Co=High temperature=2Fe+3CO2160 112800T*60% x solution x=336T

2 The mass of the caCO3 precipitate formed after the reaction is:

Conservation of mass 50g+70g=precipitation+115

Pellet = 5 g of Na2CO3 participating in the reaction.

cacl2+na2co3=caco3↓+2nacl111 106 100x 5

Solved x = mass fraction of solute in the solution after filtration (exactly.

Because it happens to react, the solute is only NaCl

cacl2+na2co3=caco3↓+2nacl111 106 100 117g

5 xx = so the mass fraction of the solute is.

1.①112÷160×100%=70%

Fe2O3 + 3Co = High temperature = 2Fe + 3CO2160T 112T800 60%T m m = 336T

2.The total mass before the reaction is 50 + 70 = 120g and the filtered solution is 115g

m=120-115=5g

cacl2+na2co3=caco3↓+2nacl106g 100g 117gn 5g q n=53g q=③

Question 1: 1Mass fraction of iron element: Fe%=56 2 (56 2+16 3)=70%.

mass = 800000000g 60% = 480 tons.

Fe2O3 + 3Co = High temperature = 2Fe + 3CO2

480 tonnes m(fe) mass of iron = 480 112 160 = 336 tonnes.

Question 2:1Obviously, the reduced mass is the precipitated mass, then m(caCO3) = 50 + 70-115 = 5g

m 5g m(naco3)=106×5/100=

5g m m(nacl)=2×

So the mass fraction of the solute is == rounded)

This can be obtained according to the formula: 56*2 (56*2+16*3 x100% according to the reaction formula to 160 grams of iron oxide to obtain 112g of iron, first calculate the mass of iron oxide, and then calculate it according to the proportion.

According to the reaction equation, we can get the overall mass reduction, which is the precipitated mass, and then find the mass of sodium carbonate according to the calcium carbonate (precipitation).

According to the reaction formula, it can be seen that the latter solution is the sodium chloride solution, and the mass of sodium chloride is calculated according to the reaction formula, and the rest is simple.

1 ① w（fe）=m（fe）/m(fe2o3)*100%=(56/160)*100%=35%

M(Fe2O3) = 800T*60%=480TBecause Fe accounts for 35% of Fe2O3, M(Fe)=480T*35%=168T is generated according to the law of conservation of mass

2 According to the law of conservation of mass m(CaCl2) + m(Na2CO3) = m(CaCO3) + m(2NaCl).

So m(CaCO3)=50g+70g-115g=5g Solution: Let the mass of Na2CO3 participating in the reaction be x

cacl2+na2co3=caco3↓+2nacl1 1

x 5g solution gives x = 5g

What is the amount of solute in solution? You didn't say yes.

Hello. (1) To make the t{hcl cao} = of standard hydrochloric acid to cao, how many grams of solute is contained in 1ml of standard hydrochloric acid.

cao + 2hcl =cacl2 + h2o56g 73

1 ml of standard hydrochloric acid contains solutes.

2) What is the above hydrochloric acid for NaOH T{HCl NaOH}?

hcl + naoh =nacl + h2ot｛hcl/naoh｝=

3) What is the purity of the sample if it is completely reacted with 1g of NaOH sample containing impurities and consumes 100ml of the above hydrochloric acid (impurities do not react with hydrochloric acid)?

hcl + naoh =nacl + h2o1g *x

x=80%

(1) Suppose: the mass of zinc sulfate is x, and the mass of hydrogen is yzn+h2so4=znso4+h2

13g x y

13g xx= 161*13g/65=

13g yy=13g*2/65=

Answer: Zinc sulfate can be prepared, and hydrogen can be produced.

2) Set: The quality of potassium chloride is x

2kclo3*****==2kcl+3o2mno2

x96 149

xx=149*

Answer: Potassium chloride is prepared.

1.Let's ask how many grams of hydrogen each!

Solution: Let ZnSO4 be x and h2 be y.

zn + h2so4 = znso4 + h265 65 + 32 + 64 1 213g x yznso4: 13g of 65 = 65 + 32 + x5x of 64 = 162 13 The answer is calculated by yourself.

h2: 13 g per 65 = y of 2 per 65

65y=2 13 The answer is your own.

Write mno2 on the reaction condition, and kcl+o2 on the bottom and bring the data in.