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Let's start with the second question, sodium chloride can't make litmus blue, which means that sodium ions can't make litmus blue, and NaOH can make litmus blue, which means that it's the hydroxide ions in Naoh that make litmus blue.
Ammonia nitrate is salt. It is known that a-g are all composed of non-metallic elements, and only one is salt, and the title itself has already explained the problem. You can't hold on to books here.
In chemistry, salt is defined as a compound composed of anions and cations. Ammonium nitrate is made up of ammonia ions (cations) and nitrate ions (anions). Ammonium nitrate differs from salts like sodium chloride in that the nitrate ions and ammonia ions in ammonium nitrate are both ions formed from charged atomic clusters, and their free charge gives them completely ionic properties (like metal ions and monoatomic anions).
And the physical and chemical properties of ammonium nitrate are similar to those of other salts.
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On the first floor, you know? Is ammonium salt called salt?
The second question is solved first.
Sodium chloride and calcium hydroxide solutions can be dropped into two litmus test solutions.
Observe the color change, which can be compared with NaOH, indicating that Na+ cannot change.
Hue. And oh- can change the color of litmus.
The first question is under consideration. Please wait, it will be solved by next Friday at the latest, 15 minutes if it is fast.
Resolved.
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a。Right. The addition of BG iron is not enough to reduce all the copper, and only a part of the copper is replaced.
b。Wrong. If there is still iron remaining, it means that all the copper in the original solution has been replaced, and the excess iron must be larger.
c。Wrong. There must be copper ions in the solution. If not, it means that the copper has been completely replaced, and the mass of C will be at least equal to A (and more likely greater than A).
d。Wrong. The oxidizing properties of 3-valent iron ions are stronger than those of copper ions. If 3-valent iron remains, there can be no solid precipitation.
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Obviously, this question is that the hydrogen emitted by the beaker A [with iron] must be more than the hydrogen emitted by the beaker B [with zinc] [in other words, the mass added by the beaker A must be smaller than the mass increased by the beaker B].
And if all the "dilute sulfuric acid containing grams of sulfuric acid" participates in the reaction, then as much hydrogen is released, so there must be a surplus of sulfuric acid in the beaker. In this way, let's consider the first possibility: too much iron, not enough zinc.
Let they all be a grams, then a beaker increases the mass =
g, b beaker increases mass = 63a65
g, the difference between the mass increase of the two beakers = 63a 65
g-(g)=, substitute a=, get.
Lower limit value = grams. [This is the minimum value of the answer, which should be greater than it, not including it.] 】。
Consider the second possibility: both iron and zinc are insufficient, and if they are all A grams [A, then A beaker increases mass = 54A 56 grams, and B beaker increases mass = 63A 65
g, the difference between the mass increase of the two beakers = 63a 65
g - (54a 56 g) =
9a 1820, substitute a=, get.
Upper limit = grams. [This is the maximum value of the answer, which should be smaller than it, not including it.] 】。So, I think the answer should be that the AB beaker mass difference before the reaction is greater than 0 and less than grams, and the mass of metal added should be greater than 0 and less than grams.
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