How to solve chemistry problem 3 I don t understand

Solution: If the mass of 5% ammonium nitrate solution is x, then the mass of 30% ammonium nitrate solution is 100kg x, which can be obtained according to the quality of the solute before and after the preparation of the solution.

5%x+(100kg x) 30%=100kg10%solution x=80kg

Then a 30% ammonium nitrate solution is required for a mass of 100kg, 80kg=20kg

100kg of 10% ammonium nitrate solution solute = 100 * 10% = 10g, then water is 100-10 = 5% ammonium nitrate solution solute 200 * 5% = 30% ammonium nitrate solution solute 50 * 30% = 15g. Assume a solution with XG 5%. YG 30% solution.

The equation x+y=100 (5%x+30%y) (x+y)=10% can be listed. If y=20, then x=100-20=80

Steps to solve the equation: (5%x+30%y) (x+y)=10%. Since x+y=100

So 5%x+30%y=10Simplify. 5% (x+6y) = 10 Divide the two sides by 5% to get x+6y=200

Since x+y=100, x+6y-(x+y)=200-100Get 5y=100So y=20

20 kg of 5% ammonium nitrate solution 30 kg of 30% ammonium nitrate solution.

1mol contains na particles, about one particle, and 1mol of an atom has a mass of m grams, that is, na, about the mass of an atom is m grams, then the mass of each atom is m na, m grams.

Memorize this. This is the formula for expansion.

Since the formula of n = 28, which is equal to the formula of co, it is enough to consider ch4 and o.

The formula quantity of ch4 is 16, the formula quantity of o is 32, and the average formula quantity is (16+32) 2=24, which is less than 28, so the volume of o must be greater than ch4, so the answer is c

c c c

c=c-c-c-c c-c-c-c

The percentage of the three types of water used divided by the total water consumption is the percentage of the three types of water used.

Agricultural Use: Industrial Use:

Domestic water: Mapping:

1. The bond only exists in the nitrogen molecule, and there is a triple bond between the nitrogen and nitrogen atoms, so there are 2 bonds in a nitrogen molecule, and the structure of N2H4 is shown in the figure, and the 4molN-H bond break is 1molN2H4 reaction to generate nitrogen, so the bond has.

2. N2H6SO4 is an ionic compound, and the structure is shown in the figure.

The third question is a small question, on the basis of the first two questions, we already know that cuprous oxide first reacts with nitric acid to form copper nitrate and elemental copper, so the elemental copper that reacts with nitric acid to form gas has a total of original elemental copper) + elemental copper formed by the reaction of cuprous oxide and nitric acid) =. The amount of its physical transport shelter = . The amount of gaseous matter produced = the mol of destruction

According to the two main chemical equations for the reaction of copper and nitric acid, the quantities of substances that form no and no2 are x and y, respectively

3cu6hno3

3cu(no3)2

2no4h2o

2cu4hno3

cu(no3)2

2no22h2o

3x/2xyy3x/2+y=

x+y=solution of the system of equations.

x=y=method introduced, check whether the specific data calculation is accurate! )