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Solubility problems in junior high school chemistry Find two questions about this, online, etc 2
Put one beaker on each side of the tray balance and adjust it to equilibrium. Inject dilute HCl with the same mass and the same mass fraction of solute into the two beakers respectively, and then put mg and Al with the same mass respectively. After the full reaction, if there is residual mg, it can be observed that
A. Keep the balance
B. Al has a surplus
C. The balance points to mg
D. The balance points to al
When equal mass of Mg, Zn and Fe are reacted with equal mass and equal solute mass fraction of hydrochloric acid respectively, only iron is completely reacted. After the reaction, the following conditions will certainly occur
A. The mass of H2 produced by the three metals is not equal to each other
B. There is a surplus of metal Mg
C. The mass of H2 produced by Zn is the least
D. There is a surplus of Zn
1. According to the equation
3Mg ＋ 6HCl ＝ 3MgCl2 ＋ 3H2
2Al ＋ 6HCl ＝ 2AlCl3 ＋ 3H2
9Mg ＋ 18HCl ＝ 9MgCl2 ＋ 9H2
8Al ＋ 24HCl ＝ 8AlCl3 ＋ 12H2
Because the mass of Mg and Al is the same, and the mass of HCl is the same, the above four equations are listed respectively, which can directly analyze the problem
From the analysis of the latter two equations, it is concluded that:
That is to say, for the same mass of Mg and Al, the acid consumed by Mg is much less than that consumed by al. It is also known that if there is a surplus of Mg and the amount of HCl added on both sides is the same, then there must be a surplus of al. Because with such a little acid, even Mg can not be completely reacted, and Al can not be completely reacted
From the analysis of the former two equations:
Because the amount of acid is the same, and all the acids react, the amount of H2 generated is the same. Because the balance is balanced when the metal is added at the beginning (the acid with the same mass, the metal with the same mass), and the mass of H2 is the same after the reaction, And there is no loss of other substances (the solid doesn't fly out of the beaker...) so the balance is balanced
So ab
－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－
2. The method is similar to 1
First of all, mgznfe reacts with acid, all of them are + 2-valent. Therefore, they react with insufficient acid of the same mass to produce the same H2
Because when the valence of metal is equal, the same mass of metal reacts with acid. The smaller the molecular weight of metal, the more acid consumed. The molecular weight of three metals is Zn > Fe > mg. It is also known that only Fe reacts completely
So Zn is not enough and acid is too much, Fe is right and acid is right, Mg is too much and acid is not enough
Therefore, the analysis of ABCD is as follows
A. Mg and Fe produce the same H2, but Zn is different from them
B. MG is excess and acid is insufficient, so mg is surplus
C. The amount of H2 produced by Zn is the least because Zn is not enough and there is residual acid after reaction
D. If Zn is not enough and the acid is surplus, then Zn must be completely removed
So choose BC
PS: detailed can not be more detailed... Points is not less
When the mass fraction of saturated solution of substance A is 20% at t ℃, the solubility of substance A is 0--------
When the solubility of substance B is 60 g at t ℃, the mass fraction of saturated solution is 0——
(1) Analysis: at t ℃, the mass fraction of saturated solution of substance A is 20%, that is, at t ℃, the saturated solution of 100g substance a contains 20g solute, 80 solvent, and the mass ratio of solute to solvent is 20:80. At t ℃, the solubility of substance A is XG, that is, 100g solvent can dissolve XG solute at most
Saturated solution = solute + solvent
100 20 80
X 100
20:80=X:100
X=25
(2) At t ℃, the solubility of substance B is 60 g, that is, at this temperature, 100 g water can dissolve 60 g solute at most, so the mass fraction of its saturated solution is as follows:
60/（60+100）*100%=37.5%
It is known that the solubility of potassium nitrate at 20 ℃ is 31.6g, and the solubility at 40 ℃ is 63.9g. Now, the 163.9g saturated solution of potassium nitrate at 40 ℃ is cooled to 20 ℃. How many grams of crystals are precipitated?
63.9-31.6 = 32.3g, the reason is that the solubility at 40 ℃ is 63.9g, and the solubility is 100g water as the solution. At this time, the maximum mass that can dissolve the solute is the solubility at that temperature, and 163.9g is 100 + 63.9 is the saturated solution with the solubility of 63.9g at 40 ℃. When the temperature drops to 20 ℃, the solute will be dissolved
At a certain temperature, evaporate 50 grams of water from 150 grams of 20% potassium nitrate solution and cool it to the original temperature. As a result, 11 grams of potassium nitrate crystals are precipitated. What is the solubility of potassium nitrate at this temperature?
Wrong answer on the first floor, 33.3%
KNO3: 150x20% = 30 water; 150-30 = 120 after evaporation of 50g water, water is 120-50 = 70g, solute is 30-11 = 19g, so the solution is saturated, so 19 / 70 = x / 100; X = 27.142g. If it is different from me, the answer is wrong
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