Chemical problems of quantity and concentration of higher substances The laboratory is equipped with 100ml of 1mol / L sulfuric acid. How many ml of 98% concentrated sulfuric acid (density of 1.84g / ml) should be taken?

Chemical problems of quantity and concentration of higher substances The laboratory is equipped with 100ml of 1mol / L sulfuric acid. How many ml of 98% concentrated sulfuric acid (density of 1.84g / ml) should be taken?

Amount of substance requiring sulfuric acid n (H2SO4)
=1mol/L*0.1mol
Its mass m (H2SO4) = 0.1mol * 98g / mol = 9.8g
It is assumed that 98% volume V (98% H2SO4) is required
=9.8g/(1.84g/mol)=5.33ml
Answer:
The laboratory is equipped with 100ml of 1mol / L sulfuric acid, and 5.33ml of 98% concentrated sulfuric acid is required

It is known that 4G hydrogen can completely react with 32g oxygen. Calculate the mass generated after 16g hydrogen reacts with 16g oxygen

If 4G hydrogen can completely react with 32g oxygen, the mass ratio of hydrogen to oxygen is 1:8
Therefore, 16g oxygen can only react with 2G hydrogen to produce 18G water

Senior one chemical calculation problem, urgent for detailed explanation! Help do it. I want to explain it in detail. Thank you Title: it is known that for a mixture of CuO and Cu2O, the mass ratio of Cu atom to O atom is 5:3. Find: (1) the mass fraction of copper atom in the mixture; (2) The ratio of the amount of CuO and Cu2O in the mixture

(1) Let Cu atom have 5mol and O atom have 3mol
The mass of Cu is 64 × 5=320g
The mass of O is 16 × 3=48g
m(Cu):m(O)=20:3
Cu%=86.96%
(2) Set cuoxmol, Cu З ohmol
have
(x+2y)/(x+y)=5:3
Solution
y=2x
That is, the ratio of CuO to Cu2O is 1:2

1. Dissolve 2.44g of BaCl · xH2O crystal in water to make 100ml solution. Take 25ml of this solution and react with AgNO3 solution containing 0.005 mol to completely precipitate chloride ions Find the amount of 2.44 g of BaCl · xH2O, the molar mass of BaCl · xH2O and the straight line of X 2. Ignite 20ml of mixed gas of hydrogen and chlorine at 150 ℃, return to the original state, and 16ml of gas remains What are the volumes of hydrogen and oxygen in the mixed gas 3. Mix sufficient concentrated hydrochloric acid and 4.35g manganese dioxide for heating, add 10.6g 10% sodium carbonate solution to the reaction solution, just no more gas is generated, and add excess silver nitrate to the solution Calculate the mass of chlorine generated and the mass of silver chloride precipitation Thank you first, little brother. Don't forget the steps. Give 30 points first, and then add if you're OK

1. N (BaCl2) = 0.005/2 = 0.0025mol in 25ml of this solution, so n (BaCl2. XH2O) = 0.0025 * 100 / 25 = 0.01mol molar mass in 2.44g of crystal, M = 2.44/0.01 = 224g / molx = (224-208) / 18 = 22, 2h2 + O2 = = gas volume before and after 2H2O reaction

At room temperature, mix 20.0 g of 14.0% NaCl solution with 30.0 g of 24.0% NaCl solution to obtain a density of 1.15g/cm ³ A mixed solution of Q: (1) what is the mass fraction of the mixed solution? (2) What is the concentration of the substance? (3) How many grams of NaCl need to be dissolved in 1000ml water to make the concentration exactly equal to the mass fraction of NaCl in the above mixed solution 20.0% 3.93mol/L 250g

Since the mass remains unchanged and the volume changes after mixing, the volume of the solution is calculated from the density
The total mass is 50 grams, the density is known, and the volume of the mixed solution can be calculated,
And the first question
Since the sodium chloride in the first solution is 2.8 g, the second solution is 7.2 G
Therefore, the total sodium chloride after mixing is 10 G
The total solution mass is 50g, so the mass fraction is 20%
As the mass of sodium chloride is 10 g, the molar number is 10 / 58.5 = 0.171
The volume is 50 / 1.15 = 43.48cm3, so the quantity concentration of the substance is 0.171 / 43.48 * 1000 = 3.93
The third equation is set to add x G
X / (x + 1000) = 20%
The solution is x = 250

The formula amount of salt m is a, and its crystalline hydrate formula amount is 2A. When t ° C (M solubility is SG), adding DGM to a certain amount of M aqueous solution just becomes a saturated solution. If its crystalline hydrate is added to reach the saturated solution, how many grams of crystalline hydrate should be added (process ~)

The mass ratio of m to water in the crystalline hydrate of M is 1:1. If the crystalline hydrate added with XG reaches saturation, the mass of m in XG crystal is 0.5x and the mass of water is 0.5x
It turns out that when DGM is added to reach saturation and replaced by crystal, the remaining m after subtracting DG from the mass of M and the water in the crystal just reach saturation, which is obtained by (0.5x-d) / 0.5x = s
X=2d/(s-1)