There are 50 chemistry calculation questions in the first year of high school, including 30 processe

According to the reaction formula, the change in the concentration of C is three times that of A, and the process is shown in the figure below.

(1) n=m m=, v=, c=n v=(2) =m m liquid =

3) c=, =, =, m=cvm=, n=cv=(4) change after dilution, because the n of the solute does not change, for example, from 10ml diluted to 100ml, the volume becomes 10 times the original, so the concentration becomes the original 1 10

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The amount of the substance of sodium hydroxide n= mol; So the concentration.

1 c==n/v== mol/l

2 c = 1000 * density * mass fraction m, so the mass fraction of sodium hydroxide = c * m 1000 * density =, a few hundredths of 20

3 or ; The mass fraction is also unchanged, 20 percent, the density, but contains less sodium hydroxide, because the volume is reduced by 10 times, the mass is, and the amount of the substance is.

4 x= mol/l

1) The molecular weight of NaOH is 40The concentration of the amount of NaOH in this solution is a = (2) and the mass fraction of NaOH is b =

3) The quantity, solubility, mass fraction and density of the substances taken out of the solution are the same as those of the original solution. The last two questions are easy to count, and I don't ...... typing

4) Diluted concentration: (10*a) (100 1000).

This problem should be solved with poor quality. NaCl (, I2 has been sublimated) was generated from Nai(150). Solids mass reduction.

Nai in the original solution: 300 g*

2Nai + Cl2 == 2NaCl + I2 solid mass reduced by 2*150g--71g---2*

--x---

-x=71*5g/183=

300g of nai has nai in it, and the equation is:

2nai + cl2 =2nacl +i2300 71 117 254― = ―

xg is solved to get x=

Answer: The equation for the reaction of copper and concentrated sulfuric acid in the question Cu + 2H2SO4 (concentrated) = Heating = CuSO4 + 2H2O + SO2 There are two reactions when adding iron powder, one is to reduce copper, and the other is to consume the remaining sulfuric acid, because there is an excess of iron Based on this idea, it can be concluded that one part of copper consumes two parts of sulfuric acid, and one part of iron consumes one part of sulfuric acid.

Analysis: annual water flow q 3600 24 365, the thermal energy of these water ep mgh vgh 103

Convert electrical energy to e ep Answer:

The precipitates of Baso4 are 1mol, 2mol and 3mol, respectively, according to SO42-conservation.

Then you need 1 3mol of aluminum sulfate, 2mol of zinc sulfate, 3mol of sodium sulfate and the same volume The concentration ratio is (1 3): 2:3=1:6:9 * I hope it can help you!