1.If 2g of oxygen gas contanins x molecules,how many molecules are present in 28g of nitrogen gas?
(Relative atomic masses:N=14.0,O=16.0)
2.If there are x molecules in 17.75g of chlorine,how many molecules are present in 8g of sulphur dioxide?
(Relative atomic masses:O=16.0,S=32,Cl=35.5)
3.Which of the following gases contains the largest number of molecules at room temperature and pressure?
(Relative atomic masses:H=1.0,C=12.0,N=14.0,O=16.0,F=19.0,Cl=35.5)
A 4.0g of hydrogen chloride
B 38.0g of fluorine
C 18.0g of ammonia
D 58.0g of carbon dioxide
4.32.18g of an oxide of metal M,when complety reduced by carbon,produced 4.95g of carbon doxide.What is the empirical formula of the oxide?
(Relative atomic masses:C=12.0,O=16.0,M=63.5)
5.4.68g of a metal X combine with 2.16g of oxygen to form an oxide in which the charge of the ion of X is +3.What is the relative atomic mass of X?
(Relative atomic mass:o=16.0)
6.The formula of hydrated sodium carbonate cystals is Na2CO3.xH20.When 66.9g of the hydrated crystals are hearted,24.8g of anhydrous anhydrous sodium carbonate are produced.What is the value of x?
(Relative atomic masses:H=1.0,C=12.0,O=16.0,Na=23.0)
(Chem) mole concept
2008-04-08 1:59 am
回答 (2)
2008-04-08 3:13 am
✔ 最佳答案
1. the no. of mole of oxygen=2/32=x moleculesthe no. of mole of nitrogen=28/28=y molecules
y=16x
2. the no. of chlorine molecule=17.75/71=x
the no. of mole of sulphur dioxide=8/64=y
y=1/8x(71/17.75)
y=x/2
3. the answer is D. the no. of mole of 58/44=1.32
4. let the formula be M2Ox
M2Ox + x/2C -------> 2M + x/2 CO2
1: x/2= [32.18/(127+16x)] : 4.95/44
after solving, x=1
the empirical formula is M2O
5. 4X + 3O2-------> 2X2O3\
4:3=4.68/X :2.16/32
after solving, X=52
6. since the molar ratio of 1:1
66.9/(106+18x) : 24.8/106
after solving X=10
2008-04-08 2:41 am
you should know and remember that ((1mole = 6.022x10^23 of anything.)) and the unit of atomic mass is [ g/mol ] =gram per mole.
why do I say you should know that? it is because once you know the unit of those stuff, you will know how to solve your homework problem in an easy way.
1.If 2g of oxygen gas contanins x molecules,how many molecules are present in 28g of nitrogen gas?
((If 2g of oxygen gas contanins x molecules))<---THIS STATEMENT IS NOT USEFUL FOR YOUR QUESTION. so the problem is just ((how many molecules are present in 28g of nitrogen gas?))
you should know the dofference between ((nitrogen gas)) ,& ((nitrogen atom))
in gas form, nitrogen appears with a form N2, which means one nitrogen molecule contains two nitrogen atom.
1. (28g)(mol/ 28.0g)( 6.022x10^23)
2. [8/(32+16x2)](6.022x10^23)
why do I say you should know that? it is because once you know the unit of those stuff, you will know how to solve your homework problem in an easy way.
1.If 2g of oxygen gas contanins x molecules,how many molecules are present in 28g of nitrogen gas?
((If 2g of oxygen gas contanins x molecules))<---THIS STATEMENT IS NOT USEFUL FOR YOUR QUESTION. so the problem is just ((how many molecules are present in 28g of nitrogen gas?))
you should know the dofference between ((nitrogen gas)) ,& ((nitrogen atom))
in gas form, nitrogen appears with a form N2, which means one nitrogen molecule contains two nitrogen atom.
1. (28g)(mol/ 28.0g)( 6.022x10^23)
2. [8/(32+16x2)](6.022x10^23)
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