A few chemistry questions, thanks!

1.The solution is electrically neutral. So the positive charge and the negative charge are equal, the positive ions have na and mg, and na has 4n, so mg has 5n, cl has 8n, and the positive charge has a total of:

4n+5n*2=14n, so s04 has: (14n-8n) 2=3n

2.The chemical reactions that occur are:

2NaHCO3=Na2CO3+CO2+H2O 2Cu+O2=2CuO62g, so the mass of CO2+H2O lost by the first reaction is equal to the mass of O2 obtained by the second reaction. Let the mass of NaHCO3 be 168g, then the lost mass is 44+18=62g, so the mass of oxygen obtained is also 62g, and the mass of Cu is calculated to be 248g, so the mass fraction of copper in the mixture=248 (248+168)=

It reacts with N2 to form Mg3N2, and reacts with water to hydrolyze. So the chemical reaction equation is:

3mg+n2==mg3n2 mg3n2+6h2o==3mg(oh)2+2nh3

It is worth mentioning that the first reaction is high temperature, and the second reaction is very violent at room temperature.

1. There are four ions of Na, Mg S04 Cl, if Na: Mg Cl = 4:5:

8 , if there are 4n na, then s04 may have 3n 4n+5n*2-8n=6n 6n 2=3n

2. When a certain amount of NaHCO3 and Cu mixture is heated at a temperature of 270° to the point that the mass before and after the reaction is found to be equal, the mass fraction of Cu in the original mixture m(H2CO3)=m(Cu) m(NaCO3)=168 62*m(H2CO3).

Mass fraction of cu = 1 (1 + 168 62) = ....

3, the product of the reaction of mg and N2 is light yellow, and the product and hydration are combined to form alkali 2cc3mg+n2==mg3n2 mg3n2+6h2o==3mg(OH)2+2NH3

The landlord hit the wrong place Sulfate is -2 valent.

1.Excuse me, in solution! If yes, the answer is 3n.

From the proportional relationship given by the title, it can be seen that if Na+=4N, then Mg2+=5N, Cl-=8N, from the principle of sensual neutrality of the solution, it can be obtained: Na+ +2mg2+ = Cl- +2SO42-, that is: 4N+2 5N=8N+2SO42-, the solution is obtained;

so42-=3n

2.The first thing to say is that the reaction device must be a reaction device with good air tightness that there is air in it, and the outside air cannot enter the reaction device 2NAHCO3 = NA2CO3 + CO2 +H2O, and the reduced part should be regarded as CO2 and H2O, and the molecular weight of the two parts is 62

The mass will be equal because the CO2 and H2O produced react with copper and oxygen in the air inside the device to form basic copper carbonate.

2Cu + O2 + CO2 + H2O = Cu2 (OH) 2CO3, because oxygen is in the original air, but the form has changed, so the increased part is also CO2 and H2O, and the mass of this part is mg by the difference method.

For the equation: 2NaHCO3 = NA2CO3 + CO2 + H2O

2 nahco3-(co2+h2o)

m（nahco3) m

Solution: m(naHCO3)=

For the equation: 2Cu + O2 + CO2 + H2O = Cu2(OH)2CO3

2cu-(co2+h2o)

m（cu） m

The solution is: m = therefore: the total mass of the mixture is:

The mass fraction of copper is: (

3.3mg+N2=mg3N2 (reaction condition is ignition).

It can be seen that Mg can react with O2 CO2 N2 in air.

mg3n2+6h2o=3mg(oh)2+2nh3↑

If you don't think about it or make a mistake, please advise and learn from each other, I hope it will be helpful to you.

3fe2o3~6fe2+~k2cr2o71m

m=mass fraction of Fe2O3=

Mass fraction of al2o3 = (

I also want to ask, I feel that the conditions are not enough.

(1) These four elements are A-C, B-O, C-Na, D-S, and (2) the chemical formula of C2D: Na2S

3) The chemical equation for the reaction between C2B2 and Ab2, and the direction and number of pips transferred.

2Na2O2 + 2CO2 = 2Na2CO3 + Direction of O Electron Transfer 2Na2O2 O Number in 2Na2O2 Lose 2E- 2Na2O2 O Number in O 2Na2CO3 2E-

c o na s

na2s2na2o2+2co2=2na2co3+o2

The direction of electron transfer is the transfer of one oxygen atom of sodium peroxide to another oxygen atom.

a, which are C, O, NA, S

Na2S2+2CO2===2+O2 O in Na2O2 gets 2 electrons reduced to O2 in Na2CO3- loses 2 electrons to zero valence of O.

Normal erythrocyte lines are suspended in isotonic plasma, and if placed in a hypertonic solution, the erythrocytes lose their internal fluid and shrink; On the contrary, if it is placed in a hypotonic solution, water enters the red blood cells, causing the disc shape of the red blood cells to become spherical on both sides, and if it continues to expand, it will rupture and dissolve, forming hemolysis. The salt concentration of plasma in the normal living environment of red blood cells is that if the concentration of surrounding salt is greater than the concentration of plasma in the living environment of red blood cells, red blood cells will lose water and shrink; When the concentration of the cell fluid is equal to the concentration of the aqueous solution around the cell, the cell retains its normal morphology; Red blood cells are a type of blood cell that lives in plasma, and the plasma solution has a concentration of 9 grams of NaCl and 1 liter of glucose.

So by calculation, 27 questions choose A, and 28 questions choose D

1. (1) I2 and H2O dispersion forces, induced forces, and (2) I2 and CCL4 dispersion forces.

3) HCl and H2O dispersion forces, orientation forces, inducing forces (4) H2O and NH3 dispersion forces, orientation forces, inducing forces, hydrogen bonding (5) O2 and H2O dispersion forces, inducing forces.

2.The trial intermolecular forces illustrate the following facts:

1) Fluorine at room temperature, chlorine is a gas, bromine is a liquid, and iodine is a solid

The molecular structure is similar, and the greater the relative molecular mass, the greater the intermolecular force and the higher the melting and boiling point.

2) The boiling point of the noble gases He, Ne, Ar, Kr, Xe increases with the increase of atomic weight.

The greater the relative molecular mass, the greater the intermolecular forces and the higher the melting and boiling point.

3) The melting and boiling points of HCl, HBR and Hi increase with the increase of molar mass.

The molecular structure is similar, and the greater the relative molecular mass, the greater the intermolecular force and the higher the melting and boiling point.

4) I2 is insoluble in water and easily soluble in CCL4.

Similar and dissolving. Both I2 and CCL4 molecules are non-polar molecules, and H2O is a polar molecule, so I2 is easily soluble in CCL4