Grade 9 chemistry questions, please answer them in detail

Left object right code. Before the balance is weighed, the pointer is to the right at the zero point, that is.

A little more object had to be placed on the left disc to center the pointer, so the object placed first was redundant.

Continuing to weigh as before, the weight mass is equal to the mass of the object minus the first point.

It is the weight mass of the object.

So the data "Actual Quality. Pick B

b is on the small side. It can be thought of in the limit method:

The pointer is right-sided before weighing, but the finger is pointed at the dial** tick mark when weighing. So we can think like this, add a certain amount of objects to the left disk, so that the finger is aimed at the index dial ** scale, at this time the right disk has not added weights or moved the code, so the reading is 0, but at this time the mass of the object is no longer 0, so the reading is small.

b is on the small side. As standard, if the pointer is right, the balance is balanced without the actual weight of the weight.

The balance hand is deflected to the right, and the right weighing pan is considered to have a greater mass than the left (this is hypothetical, but actually equal, and is called equivalent substitution in physics). Then according to the weighing method of the left object and the right code, the mass of the object = weight mass + right weighing pan mass - left weighing pan quality, if the weight mass is the experimental data, the actual mass is greater than the experimental data, so B is selected

b. Put the object on the left and the weight on the right. Initially to the right, part of the object on the left to balance the balance, so the weighing is larger than it actually is.

When weighing the balance, the left and right weights the object. Before weighing, the pointer is biased to the right, which is equivalent to a small weight (δm) has been added. When weighing, when the balance reaches equilibrium, the mass of the item should be the weight of the weight (the data obtained from weighing) plus δm. Therefore, the obtained data is small compared to the actual quality, so B is selected

The water in the beaker flows backwards into the jar, which accounts for about 1 5 of the jar volume

Phenomenon: Water from the beaker flows backwards into the jar, accounting for about 1 5 of the volume of the jar

Analysis: The combustion of red phosphorus consumes the oxygen in the bottle, so that the pressure inside the bottle is reduced. When the clip is opened, water is drawn into the bottle. until it is just the size of the volume of oxygen consumed in the bottle. About 1 5,

Note: Wait for the jar to cool to room temperature before opening the water stop.

Phenomenon: Water from the beaker enters the jar along the catheter, and the volume of water entering the bottle is approximately 1 5

Analysis: The combustion of red phosphorus consumes the oxygen in the bottle, which reduces the overall volume of gas in the bottle, thereby reducing the air pressure in the bottle. It is less than the outside atmospheric pressure, so the atmospheric pressure presses the water into the bottle.

P+O2--P2O5, phosphorus combustion consumes oxygen, so, the oxygen in the gas collector cylinder is consumed, the air pressure is lower than the outside atmospheric pressure, when the clamp is released, the external atmospheric pressure will press the air of the beaker into the gas collector cylinder, and the water surface in the gas collector cylinder rises.

A (grinding produces (NH4OH will eventually volatilize ammonia) BC (both form two precipitates insoluble in water and acid and alkali) D (the most special SO2, which has similar chemical properties to CO2 and can also make clarified lime water turbid).

So choose D

There is also a reaction between bicarbonate and acid to produce carbon dioxide, which makes the clarified lime water turbid.

For example, sodium bicarbonate, NaHCO3, and hydrochloric acid also react to produce carbon dioxide. ......

d, sulfite ions are also in line with the sulfur dioxide gas produced by sulfite in hydrochloric acid, which can also be clarified lime water turbidity.

It doesn't have to be the salt of CO32, the salt of HCO3 is okay too.

The gas released by the reaction of sulfite and hydrochloric acid is sulfur dioxide, which can also make the clarified lime water turbid and generate calcium sulfite precipitation.

Chemical Equation:

na2so3 + 2hcl == 2nacl + so2↑ +h2o

so2 + ca(oh)2 == caso3↓ +h2o

This kind of question is generally to look at the change in the mass of the substance or the amount of the substance before and after the reaction to determine which substance in the reaction is a reactant or a product or a catalyst. We know that the mass of reactants decreases after reaction, and the mass of products increases, and the mass of catalysts generally remains unchanged, which excludes the case of catalyst poisoning in complex reactions.

So, analyzing this question, the mass of M before and after the reaction is 2, so it may be a catalyst. If the mass of n increases, then it is a product; If the mass of x decreases, then it is a reactant, and if the mass of y increases, then it is a product.

x=n+y, the mass change of n is 18g, y is 16g, so the mass ratio is 9:8, so the reaction can be obtained as a decomposition reaction, and B error is obtained.

First of all, the mass of the reactant decreases, the mass of the product increases, and the mass does not change, which may be a catalyst or may not be related to the reaction.

This is the method and law of doing this kind of question.

n by 20---38, which is the product, x by 34---0, which is the reactant.

y is made up of 0---16, which is a spawn.

Therefore, the reaction is: the mass ratio of x---n+ and y is: 18:16=9:8, and answer b is false.

Select B first calculates the mass change of each substance before and after the reaction, the catalyst with the same mass, the product with an increase in mass, and the reactant with a decrease in mass.

m mass unchanged, is the catalyst; n The mass of hydroxide increased by 18g, which is the product; The mass of X was reduced by 34g after the reaction, which was a reactant; The Y mass has increased by 16g and is a product.

So this reaction can be written as: x=n+y, and the reaction condition is m catalysis, which is a decomposition reaction with a mass ratio of n to y of 18:16=9:8

ACD is correct, B is false.

b, a chemical reaction occurs, the mass before the reaction should be equal to the mass after the reaction, and the elements in it will not decrease, nor increase, but only transform from one form to another. After the reaction, n increased by 18 and y increased by 16, so the mass ratio was 9:8.

And X is gone, and the mass of M has not changed, then it means that X is a reactant and has decomposed, N does not participate in the reaction, it is a product, Y is also a product, and the mass before and after M has not changed, which means that it has not participated in the reaction and plays a catalytic role, indicating that M is a catalyst. It's clear that this is a decomposition reaction.

A: NaCl has a valency of -1

B: HCO4 has a valency of +7

C: Cl2 elemental, valency of 0

d: ca(clo)3 (it seems that there is no such substance, it should be ca(clo)2, then the valency is +1.)

So the lowest valency is A