One chemistry calculation problem for the third year of junior high school, and one chemical calcula

m dilute sulfuric acid = density * volume = cubic centimeter * 500 cubic centimeter = 535gm sulfuric acid = m liquid * p% = 10% * 535g =

m concentrated sulfuric acid = m sulfuric acid p% =

v concentrated sulfuric acid = m concentrated sulfuric acid density = cubic centimeter cubic centimeter = m water = m dilute sulfuric acid - m concentrated sulfuric acid =

v water = mass density = cubic centimeter =

Answer: The mass of 500ml of 10% dilute sulfuric acid is 535g, and 98% concentrated sulfuric acid is needed, that is, water.

Requires 98% concentrated sulfuric acid) =

It is concluded that 98% concentrated sulfuric acid water is required =

The total mass of dilute sulfuric acid after preparation is 500

Set the volume of concentrated sulfuric acid x (unit ml) required

x solution x 2967ml [3L].

Its mass is 2967

Hence the quality of the water is.

Volume psThe second step upstairs should be the volume (ml) instead of the mass (g), otherwise the density would not be used in the front

The mass of dilute sulfuric acid is 500ml * cubic centimeter = 535g The mass of concentrated sulfuric acid is 500ml * cubic centimeter

i.e. cubic centimeters =

Water requirement (cubic centimeter =

m = v, so m = cubic centimeter 500ml = 535g because m concentrated % = m dilute %, so m concentrated = 535g 10% 98% v concentrated = cubic centimeter cubic centimeter cubic centimeter can be obtained by the formula of 1.

m plus water = m dilute - m concentrated =

The calculations I have kept 2 decimal places, and you can take the number of digits you want to keep as many digits as you want.

Ammonium nitrate molecular formula: NH4NO3

Nitrogen content: n%=28 80=35%.

Product purity = 34% 35% =

Mass of pure ammonium nitrate in the bag = gross weight Purity = 50 The mass of pure ammonium nitrate in the bag is calculated as kilograms.

Some of the oxidized Fe sheets are Fe and the other is Fe2O3. Therefore, the mass of Fe can be X, then the mass of Fe2O3 is (10-X), and the mass of H2SO4 reacting with Fe and Fe2O3 is y and Z, respectively.

The reactions that occur are:

1. Fe+H2SO4=FeSO4+H2 (with arrows, mark yourself) 56 98 2x y

56 x=98 x=2 obtain: x=,y=,10-x=2, fe2o3+3h2so4=fe2(so4)3+3h2o160 294

Z160 obtains: Z=So the solute H2SO4 contained in 200GH2SO4 is, then W(H2SO4)=

There is nothing wrong with guaranteeing the process, but the result is still up to you.

Let the mass of sulfuric acid be x

fe+h2so4=feso4+h2

98 2x98/x=2/

x = A.......

Solution: Fe+H2SO4=FeSO4+H2

Because of the production of grams of hydrogen, the mass of iron is:

The mass of sulfuric acid involved in the reaction is:

The mass of iron oxide is:

Fe2O3+3H2SO4=Fe2(SO4)3+3H2OSo the mass of sulfuric acid reacting with iron oxide is:

So the total mass of sulfuric acid is.

The mass fraction of sulfuric acid is:

Answer: The mass fraction of sulfuric acid is.

First of all, the whole process contains two chemical equations: one is the reaction of dilute sulfuric acid with iron, and the other is the reaction of dilute sulfuric acid with iron oxide.

From the inscription, we can see that the production of grams of hydrogen gas, that is, iron and dilute sulfuric acid participate in the reaction (list the chemical equation) Therefore, the oxidized iron sheet has grams of iron, grams of iron oxide. Listing the chemical equation of iron oxide and dilute sulfuric acid can be calculated to be about the amount of dilute sulfuric acid consumed by iron oxide.

In summary, the total consumption of dilute sulfuric acid is, and then the mass of dilute sulfuric acid is calculated as the solute mass fraction of sulfuric acid is.

1. Let the mass of sulfuric acid be x, and the mass of fe be y

fe+h2so4=feso4+h2

56 98 2y x

98/x=2/ x= g

56 y=2 y= g Fe is oxidized to Fe2O3 in the amount of g

2. Set the quality of sulfuric acid to k

fe2o3+3h2so4=fe2(so4)2+3h2o160 294

k160/ k= g

3. Fe and Fe2O3 share the mass of H2SO4 is g, and the mass fraction of sulfuric acid solution is.

That is, yes. Then according to 2H Fe there is oxidation, that is, it.

So the oxidized part is. Natural oxidation is said by fe2o3. So yes, it's iron, that is to say, there is still H+ that has turned into water. So sulfuric acid is.

So v

Urea chemical formula: CO(NH)2

Therefore, the farmer bought n elements (28, 58) * 500, about = 250, an average of 5 kg per mu.

co(nh2)2

m1=1000/1080*540=500kgm2=500/50=10kg

m3 = 10 * [(14 * 2) (12 + 16 + 14 * 2 + 2 * 2 * 2)] = nitrogen per mu.

The reaction chemistry equation is:

cuo+h2=cu+h2o

The total mass of the original mixture of x is 6g, after the reaction, only cu is left after the reaction, and the reduced mass is the mass of element o, that is, the mass of o, then the mass of the water produced after the reaction can be calculated as, and the mass of cuo participating in the reaction is x, then x=4g can be calculated, then cuo is 4g, and cu is 6-4=2g

Because it is a complete reaction, all the oxygen elements in copper oxide are used to generate water, so there is less oxygen in copper oxide, and because the mass fraction of oxygen in copper oxide is 16 (16+64)=, there is copper oxide.

The rest is copper, there are 6-4 = 2g

Pure zn mass m =

Reaction equation: Zn + H2SO4 ==== ZnSO4 + H2

Reaction ratio 64:98:2

6..498 ： x ： y

64:98=:x x= x is the mass of pure sulfuric acid, then the mass of dilute sulfuric acid m=

64:2=:y y= y is the mass of hydrogen.

zn = approximately, so to generate hydrogen, sulfuric acid needs at least 98g

Because: 25GA reacts exactly with 10GB to produce 5GC. So because of the conservation of mass, it can be concluded that 30 gd of matter is generated.

From this it is possible to trim the chemical formula: 5a+2b=c+6d.

So when there is 6 GD generated, there is 5 GA to participate in the reaction.

According to the mass balance: when the complete reaction is produced d 30 g, when d 6 g, the reaction is only carried out by one-fifth, so the reaction of a 25 5 g is 5 g.

(1) Co-generation of carbon dioxide gas: The density of carbon dioxide under this condition is so that the CO2 mass is m= = g

2) Let the mass of calcium carbonate in the limestone sample be x g, CaCO3+2HCl====CaCl2+H2O+CO2.100. .

44x100:44=

The mass fraction of calcium carbonate in the x=5g limestone sample is =5g

Solution: (1) Mass of carbon dioxide =

2) Let the mass of calcium carbonate be x

caco3+2hcl=cacl2+co2+h2o100 44

x100/x=44/ x=5g

Mass fraction of calcium carbonate = 5g

Answer: The mass fraction of calcium carbonate is:

1 16/32*100%=50%

2. 500t of industrial wastewater containing methanol containing 500t of methanol * 5CH3OH + 12O2 + 6NH3 = bacteria = 3N2 + 5CO2 + 19H2O160 102X

160/x=

(1) Mass fraction = 16 (16 + 12 + 4) * 100% = 50%.

2) The mass of ammonia mass of methanol = 6 * 17 5 * 32 = 51 80

So the mass of ammonia = 500t*32%*51/80=

Melamine chemical formula C3H6N6, phase Bi Shi pure pair molecular mass is 12 * 3 + 6 * 1 + 14 * 6 = 126

The mass fraction of nitrogen in melamine = 84 126 = 1g The mass of nitrogen in melamine is 1*

Quite guess the quality of the protein hand ordered to do.