There is a chemistry problem that anyone solves, a chemistry problem, I don t know

Let the hydrocarbon be CXHY

cxhy + x+y/4) o2 *****x co2 + y/2 h2o

Because the volume has been reduced by 3

So 1+ (x+y 4) -x=3

So y=8 is a gaseous hydrocarbon again, so the answer is c3h8

The equation can be written.

C(z)H(4Y-4Z)+(Y)O2==(Z)CO2(gas)+(2Y-2Z)H2O(liquid).

According to the title. 1+y=z+3*****==>y-z=2 is c(z)h8

z can be 3 or 4

c3h8c4h8

Hydrocarbon (CXHY) + (X+Y4)O2 XCo2+Y2H2O (1+X+Y4-X-Y2=1-Y4).

So 1-y 4 = 3 y = 8 x = 3,4 (because it will be gaseous below four carbons).

C3H8 (saturated hydrocarbons) or C4H8 (unsaturated hydrocarbons).

Let the molecular formula of the hydrocarbon be CXHY

Then cxhy + x+y 4) o2 == x co2 + y 2 h2o

The volume of water at room temperature is not considered.

So 1+ (x+y 4) -x=3

The solution is y = 8

So the hydrocarbon may be C3H8 C4H8 (Hydrocarbons with a number of C atoms greater than or equal to 4 are liquid or gaseous).

Question 1: The condition under which iron rusts.

In the textbook experiment, it takes a long time for iron to rust, so an experimental team designed the following improvement experiment.

1) Check the air tightness of the device: connect the device, close the piston of the separating funnel, connect the C catheter to a beaker filled with water, and slightly heat the A device, indicating that the phenomenon of good air tightness is: there are bubbles escaping from the C catheter, stop heating, and the C catheter will form a water column.

Because the volume of the gas expands when heated, more than the gas will escape, and the heating will stop, the gas will rot and shrink, and the water will enter the conduit to form a water column.

2) After the inversion of the ballast should begin, the reaction phenomenon in device A is: there is a large number of bubbles generated. 2H2O2 = 2H2O + O2 (MnO2 catalysis). MNO2 is the catalyst.

3) After 4 minutes, the wire is still bright, the surface of the wire is gray at B, and the wire is still bright at D. This experiment shows that iron rust is mainly related to H2O and O2.

The comparison of experimental phenomena at B and D shows that an important factor determining the speed of iron rusting is the concentration of O2. Because the higher the concentration of reactants, the faster the reaction rate.

The reduced sulfur element is grams, i.e., when it is reduced, the valency decreases, and K2S is generated

i.e., the s of the reaction is.

The koh of the reaction is, the concentration.

3 mols reaction, 2 mols is reduced, and 1 mols is oxidized.

2s (reduction) 3s (total) 6koh

64g 6mol so c(koh)=

Copy the self-weight.

The potassium sulfide generated is reduced, so the relationship between the reduced sulfur and the potassium hydroxide consumed is 2:6=1:3. So the potassium hydroxide consumed is (,c=

2 mols of the 3 mols reaction to form potassium sulfide is reduced, and 1 mols of potassium sulfate is oxidized.

2s (reduction) 3s (total) 6koh

64g 6mol so c(koh)=

The water contains H, contains C, and C+H is total. then o

Remind the idea, only contain the element C, there may be element O, know the quality of water, know the mass of H, find the mass of C, and compare the two with the mass of organic matter to get the mass of O.

The generation of CO2 and H2O indicates that the raw material contains C, and the ratio of H is 1:4

Heavy <, indicating the presence of the O element, ie.

So ch4o3

(1) In the thermochemical equation, the coefficient of N2 is 1, and the coefficient of H2 is 3, which means that the energy released by reacting 1molN2 and 3molH2 (is the reaction, not the input).

And the amount of input in the question does not refer to how much is reflected. So to be less than (2)n2 conversion rate, your understanding is correct. Option d is wrong.

The answer is a

The meaning expressed by the thermochemical equation is that 1molN2 and 3mol H2 completely react to form 2mol NH3, and the heat released is , and the middle 3molH2 cannot be reacted, so it is <

d is wrong and should: