In order to judge whether a compound contains carbon, hydrogen and oxygen, a certain mass of the compound should be fully burned in oxygen. The following experiments are needed: (1) using anhydrous copper sulfate to check whether there is water formation; (2) using clear lime water to check whether there is carbon dioxide formation; (3) using wood strips with Mars to check oxygen; (4) determining the mass of water and carbon dioxide A. ①②③B. ①②④C. ②③④D. ①②③④

In order to judge whether a compound contains carbon, hydrogen and oxygen, a certain mass of the compound should be fully burned in oxygen. The following experiments are needed: (1) using anhydrous copper sulfate to check whether there is water formation; (2) using clear lime water to check whether there is carbon dioxide formation; (3) using wood strips with Mars to check oxygen; (4) determining the mass of water and carbon dioxide A. ①②③B. ①②④C. ②③④D. ①②③④

Copper sulfate is a white powder. When it meets with water, it will generate copper sulfate pentahydrate and turn blue, which indicates that water is generated, that is, the compound contains hydrogen element. When the lime water becomes turbid, it indicates that carbon dioxide exists, that is, the compound contains carbon element. The judgment of oxygen element should be based on the quality. The quality of water and carbon dioxide should be determined, and the water quality should be calculated according to the quality of water The mass of hydrogen in the compound is the mass of hydrogen in the compound. According to the mass of carbon dioxide, the mass of carbon in carbon dioxide is calculated, which is the mass of carbon in the compound. The mass of carbon plus the mass of hydrogen is compared with the mass of the compound. If it is equal, there is no oxygen in the compound. If it is less than the mass of the compound, there is oxygen in the compound So B

When an organic substance is completely burned to produce carbon dioxide and water, the mass fraction of carbon element in the organic substance is 40%. Is it possible that the substance is hydrocarbon,
Please also determine the molecular formula of the substance with the smallest relative molecular weight that meets the known conditions?

No way,
CH2O(HCHO)

Both Party A and Party B deposit X Yuan in the bank. Later, Party A deposits 800 yuan and Party B withdraws 1200 yuan. As a result, Party A's deposit is three times as much as Party B's deposit. How to make an equation
Explain the equivalent relation

X+800=(X-1200)*3
X=2200

Wang Fang's deposit is 2.2 times that of Lily's. If Li Li deposits another 75 yuan in the bank, the two people's deposits will be equal. How many yuan did they each deposit

If lily's deposit is x yuan, Wang Fang's deposit is 2.2x yuan
2.2X=X+75
1.2X=75
X=75/1.2
X=375/6
So Lily's deposit is 375 / 6 yuan
2.2X=825/6
That is, Wang Fang's deposit is 825 / 6 yuan
A: Lily's deposit is 375 / 6 yuan and Wang Fang's is 825 / 6 yuan

The money ratio of a and B was 6:5, then a got 180 yuan and B got 30 yuan. At this time, the money ratio of a and B was 18:11
The original money ratio of a and B was 6:5. Later, a got 180 yuan and B got 30 yuan. At this time, the money ratio of a and B was 18:11. How much money did they have?
No equations!

Let a have 6 x yuan, then B have 5 x yuan
(6x+180):(5x+30)=18:11
11(6x+180)=18(5x+30)
The solution is: x = 60
60 × 6 = 360 yuan 60 × 5 = 300 yuan
Answer:
(180 × 11-30 × 18) / (18 × 5-11 × 6) = 60 yuan
60 × 6 = 360 yuan; 60 × 5 = 300 yuan
Do you understand!

The ratio of money between a and B was 6:5. Later, a got 180 yuan, and already got 30 yuan. At this time, the ratio of money between a and B was 18:11
How much did they have in total
No equations

Let X / y = 6 / 5 (x + 180) / (y + 30) = 18 / 11, that is, x = (6 / 5) Y11 (x + 180) = 18 (y + 30) (11 * 6 / 5) y + 11 * 180 = 18y + 18 * 30 (11 * 6 / 5) y-18y = 18 * 30-11 * 180 (11 * 6 / 5) y-18y = 18 * 30-11 * 180y = (1980-540) * 5 / (90-66) y = 1440 * 5 / 24

The original money ratio of Party A and Party B was 6:5. Later, Party A got 1800 yuan and Party B got 300 yuan. At this time, the money ratio of Party A and Party B was 18:11. How many yuan did Party A and Party B have?

Suppose that a has 6 x yuan
(6x+1800)×11=18×(5x+300)
66x+19800=90x+5400
90x-66x=19800-5400
24x=14400
x=600
6x=3600
5x=3000
A had 3600 yuan, B had 3000 yuan

The ratio of money between a and B was 6:5, then a got 1800 yuan, and B got 300 yuan. At this time, the ratio of money between a and B was 18:11, and the ratio of money between a and B was 18:11

A has 3600, B has 3000

Party A and Party B each have a certain amount of RMB. Party A takes out 1 / 5 of his own to Party B, and Party B takes out 1 / 4 to Party A. at this time, they each have 90 RMB. How much are they
Solving linear equation with one variable

In the end, if both of them have 90 yuan, they will have a total of 180 yuan
According to the meaning of the title, 4 / 5x + 1 / 4 (1 / 5x + 180-x) = 90
3/4(1/5x+180-x)=90
The solution is x = 75 and B is 105

Party A and Party B have a total of 440 yuan, 1 / 2 of the money of Party A is equal to 3 / 5 of the money of Party B. how much money do Party A and Party B have?
Please give us the specific process

Let a x yuan, B 440-x yuan
1/2x=(440-x)×3/5
1/2x=440×3/5-3/5x
1/2x+3/5x=264
5/10x+6/10x=264
11/10x=264
x=240
B: 440-240 = 200 yuan