High School Chemistry Questions, Please Help, Help How to do this high school chemistry questions?

1. There are 10 Cl-1 that lose 10 electrons to generate 10 Cl, and there are 2 Mn+7 ions to get 5 electrons to generate 2 Mn+2 ions.

2. I don't understand that i2 is an iodine molecule, how can the valency be 5?

3. Sodium and salt water can react Specifically, the reaction between sodium and water has nothing to do with NaCl. The ionic equation is 2Na+2H20=2Na+ +2Oh- +H2

4. I provide the solution to this problem: the limit method assumes that all the substances here are NaHCO3 or KhCO3. Calculate the range of gases formed by the reaction with hydrochloric acid separately (note whether hydrochloric acid is not enough).

5. Any gas of 1mol under standard conditions is.

This gives you the ability to calculate the density of N2 and CH4.

Suppose the amount of N2 in 1L gas mixture is X and the amount of CH4 is 1-X

Density of N2 * x + density of CH4 * (1-x) =

Just solve it. 6. The gas generated is H2 and O2, and the ratio of material to quantity is 2:1

You can see the quality of their things.

H2 is the reaction of Na and O2 is the reaction of Na2O2 to produce the mass of the matter of both. Na2O, on the other hand, reacts with water without forming gas.

1, look at the valency, MN changes from positive +7 to +1, and there are two molecules, which shifts 122, which is incomprehensible.

3. The essence of the reaction is to react with water. 2na+2h2o =2na^+ 2oh- +h2

4. No matter how it is mixed, v=n*l

5, let the volume ratio of N2 be x%, 28x+16(1-x)=x=20%6, according to certain conditions, the gas happens to be completely reactive, the volume ratio of H2 O2 in the mixed gas is 2:1 Let the mass of H2 be X, the mass of O2 is 7 36 x divided by the molar mass, multiplied by 22 4 equals 2:1 and it is obtained.

a false, the number of neutrons is unequal;

b right. O2+ and Na+ in Na2O are 1:2;In Na2O2, peroxide ions and sodium ions are also 1:2;Therefore, no matter how you mix it, it will always be 1:2.

c false, there is also cl2 in the water, this part of the chlorine atom is not considered;

d false, there is no standard condition.

Awesome answer:

d can only be iron.

At room temperature, Fe is passivated in concentrated H2SO4, so there is no obvious phenomenon when heating: 2Fe 6H2SO4 (concentrated) Fe2(SO4)3 3SO2 6H2O

H is Fe2(SO4)3.

fe2（so4）3＋2h2o＋so2＝2feso4＋2h2so4

D is the reaction of Fe and concentrated sulfuric acid to form three iron ions Fe2 (SO4)3 After the three iron ions oxidize the sulfurous acid to form sulfuric acid and FeSO4, I hope to help you,,

If you don't understand, you can ask 、、、

D is iron, E is iron oxide, and H is ferrous sulfate.

This problem can be solved by the conservation of electric charge.

Iron finally becomes an iron tripositive ion, that is, the lost electron electrons, and finally the n positive pentavalent atom gets into no, and a nitrogen atom needs three electrons, so it can get no

The last thing left is NO, regardless of the process, which is equivalent to the oxygen in HNO3 to Fe2O3, that is, a part of HNO3 is decomposed into H2O O NO

Fe2O3 has oxygen in Mo.

2hno3~~~h2o+3o + 2no

xx= So there are 2240ml

i.e. conservation of elements.

1) At 1026 degrees, 4cuo = 2cu2o+o2

Let the amount of Cu2O produced be x (mol).

Then, the amount of cuo participating in the reaction is 2x mol

The amount of substance added to cuo is: 8 80 = mol

So, the amount of substance left after the reaction of cuo is ( mol

It is known that in the measured post-reaction mixture, n(cuo):n(cu2o) = 1:2

i.e.: (get x= mol.)

That is, the amount of substance that produces Cu2O is mol

2) At 1026 degrees, 4cuo = 2cu2o+o2

At 1800 degrees, 2Cu2O = 4Cu+O2

Suppose the amount of Cuo in the remaining mixture is y mol, then Cu2O has 2y mol and Cu has 5y mol in the mixture

From the reaction at 1800, the amount of Cu2O participating in the reaction is 5y 2 mol, and the amount of O2 produced is 5y 4 mol

So, the total amount of Cu2O produced in the reaction is: 2y+(5y2)=9y2

Then it is known from the reaction of 1026:

The amount of Cuo participating in the reaction is 9y mol, and the amount of O2 produced is 9y 4 mol

Therefore, the total amount of Cuo added is 9y+y=10y, and the amount of O2 formed in the reaction is (5y4)+(9y4)=14y4=7y2

The quality of the added cuo is known to be 8g

So, 10y=8 80=

So, y=then, the amount of matter that is co-generated twice with O2 is 7y2=

Its quality is:

. co + h2o = co2 + h2

Start 0 0 reaction x x x x

Equilibrium x xc(co2)*c(h2) c(co)*c(h2o)=(x=?CO conversion rate under this condition = x ?

Calculate the value of x yourself, and the solution is like this.

The equilibrium constant k = the concentration of the product and the concentration of the reactant.

With a three-line formula. Let the concentration of CO2 at the equilibrium of the reaction be A, then the concentration of H2 is also ACO + H2O(G) = CO2 + H2Before the reaction.

Reaction a a a a

Post-reaction a a

k=(a*a)/(

Solve A and then use A to solve the conversion rate!

A is easy to judge is correct, because there is a change in the volume of the gas before and after, so it cannot be pressure, it can only be a catalyst!

The b reaction is an exothermic reaction, reducing the temperature, the equilibrium moves in the positive direction, and the reaction rate decreases, but it should be v positive and v inverse, so it is wrong!

c increases the pressure, and the equilibrium moves in the direction of volume reduction, that is, it moves in the positive direction, and the reaction rate increases, but it should be v positive and v inverse, so it is wrong!

d The conversion rate is the smallest, you should find the reverse movement, no reverse to find no movement, no no movement, find a small positive movement! So wrong!

To sum up, choose A!

The above analysis is correct, so I won't repeat it.

a, b to reduce the pressure.

c. Raise the temperature.

The lowest conversion rate is T1-T2

When the positive catalyst is added at T2, the forward and reverse reaction rates increase at the same time.

t3 is the decreasing pressure, the inverse is greater than the positive, and the total velocity is decreasing;

t5 is the increase in temperature, reverse endothermy, inversion is greater than positive, and the total rate increases;

T3 to T6 has been inverse and greater than positive The equilibrium shifts to the left, so the conversion rate is the lowest after T6;

In short, the catalyst, pressure (with gas), and heating rate will increase, and it is enough to consider which side the equilibrium will move.