Hello everyone! I've got some chemistry questions that I don't know how to do. I have the correct answers. Can anyone please explain briefly to me. Thank you for your help.
1. The formula for ozone is O3. If one mole of ozone contains x atoms, how many atoms will one mole of oxygen gas contain?
My problem: The answer is 2x/3. Why? How to get this answer?
2. 7.5g of calcium carbonate is added to 50.0 cm3 of 2M hydrochloric acid. What is the colume of carbon dioxide liberated at room temperature and pressure? (relative atomic masses: C=12.0, O=16.0, Ca=40.0 ; molar volume of gas ar r.t.p. =24.0 dm3)
My problem: The answer is 1.8 dm3. Why should I use the mole of hydrochloric acid for caculation but not calcium carbonate?
3. A sample of concentrated sulphuric acid has a density of 1.83gcm-3 and contains 94.0% of sulphuric acid by mass. What is the concentration of sulphuric acid in the sample?
My problem: I don't know how to do it. The answer is 17.5M. Could you please show me the caculation?
4. A mixture consists of one mole of sudium carbonate and one mole of sodium hydrogencarbonate. What is the least number of moles of hydrocholric acid required to liberate all the availble carbon dioxide from the mixture?
My problem: I have no idea. The answer is 3.0. Could you explain to me?
5. Gases X and Y react to give a gaseous product Z. The reaction can be represented by the equation: X(g) + 3Y(g) >2Z(g)
In an experiment, 4. cm3 of X and 60 cm3 of Y are mixed and are allowed to react in a closed vessel. What is the volume of the resultant gaseous mixture in dm3?
A.40 B.60 C.80 D.100
My problem: I have no idea. The answer is B. Could you explain to me?
I hope you can help me as much as the questions you can. Thanks again!
some chemistry questions
2009-08-18 9:42 pm
回答 (2)
2009-08-19 2:21 am
✔ 最佳答案
1.Avogadro constant = L /mol
(1 mol of molecules = L molecules)
Consider 1 mole of O3:
No. of moles of O3 = 1 mol
No. of O3 molecules = L molecules
No. of O atoms = 3L atoms = x atoms
Hence, L = x/3 ...... (*)
Consider 1 mole of O2:
No. of moles of O2 = 1 mol
No. of O2 molecules = L molecules
No. of O atoms = 2L = 2(x/3) = 2x/3 atoms
2.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Mole ratio CaCO3 : HCl : CO2 = 1 : 2 : 1
Molar mass of CaCO3 = 40 + 12 + 16x3 = 100 g/mol
No. of moles of CaCO3 added = 7.5/100 = 0.075 mol
No. of moles of HCl added = 2 x (50/1000) = 0.1 mol
When all HCl is reacted, CaCO3 needed = 0.1 x (1/2) = 0.05 mol
Hence, CaCO3 is in excess and HCl is completely reacted.
No. of moles of HCl reacted = 0.1 mol
No. of moles of CO2 formed = 0.1 x (1/2) = 0.05 mol
Volume of CO2 formed = 0.05 x 24 = 1.2 dm3
(The answer is NOT 1.8 dm3.)
3.
Consider 1 dm3 (1000 cm3) of the solution.
Mass of the solution = 1.83 x 1000 = 1830 g
Mass of H2SO4 in the solution = 1830 x 94% = 1720 g
Molar mass of H2SO4 = 1x2 + 32.1 + 16x4 = 98.1 g/mol
No. of moles of H2SO4 = 1720/98.1 = 17.5 mol
Molarity of H2SO4 = 17.5/1 = 17.5 M
4.
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Mole ratio Na2CO3 = 1 : 2
No. of moles of Na2CO3 = 1 mol
No. of moles of HCl = 1 x 2 = 2 mol
NaHCO3 + HCl → NaCl + H2O + CO2
Mole ratio NaHCO3 : HCl = 1 : 1
No. of moles of NaHCO3 = 1 mol
No. of moles of HCl = 1 mol
Least number of moles of HCl required = 2 + 1 = 3 mol
5. The answer is B.
(The volume of X should be 40 cm3, but NOT 4. cm3)
X(g) + 3Y(g) → 2Z(g)
Volume ratio X : Y : Z = 1 : 3 : 2
Volume of X added = 40 cm3
Volume of Y added = 60 cm3
If all Y is reacted, volume of X reacted = 60 x (1/3) = 20 cm3
Hence, X is in excess, and Y is the limiting reactant (completely reacted).
Volume of Y reacted = 60 cm3
Volume of Y left = 0 cm3
Volume of X reacted = 60 x (1/3) = 20 cm3
Volume of X left = 40 - 20 = 20 cm3
Volume of Z formed = 60 x (2/3) = 40 cm3
Volume of the resultant gaseous mixture
= Volume of X left + Volume of Z formed
= 20 + 40
= 60 cm3
2009-08-18 18:22:01 補充：
貼出問題，點數己付出，刪除題目只會損人不利己。
2009-08-20 15:26:35 補充：
答冷風，約10分鐘。
2009-08-20 3:26 am
老子好野, 用左幾多時間答?
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