1.on strong heating, 10g of hydrated sodium sulphate produces 5.6g of water. What is the mole ratio between sodium ion and water in this compound?
ans: 1:5
2.The number of atoms present in 15cm^3 methanoic acid is?
(density of methanoic acid=1.22gcm^-3)
ans: 1.2X10^24
how to calculate?
actually i am not sure if the answers given are correct or not.
Thank you!
f.5 chem
2009-04-07 5:09 am
回答 (3)
2009-04-07 7:32 am
✔ 最佳答案
1.Molar mass of Na2SO4 = 23x2 + 32.1 + 16x4 = 142.1 g/mol
Molar mass of H2O = 1x2 + 16 = 18 g/mol
Mole ratio Na2SO4 : H2O
= (10 - 5.6)/142.1 : 5.6/18
= 0.0310 : 0.311
= 1 : 10
Each mol of Na2SO4 contains 2 mol of Na+.
Mole ratio Na+ : H2O
= 1x2 : 10
= 1 : 5
2.
Molar mass of HCOOH = 1x2 + 12 + 16x2 = 46 g/mol
Mass of HCOOH = volume x density = 15 x 1.22 = 18.3 g
No. of moles of HCOOH = 18.3/46 = 0.398 mol
No. of HCOOH molecules = 0.398 x (6.02 x 1023)
Each HCOOH molecule contains 5 atoms.
No. of atoms = [0.398 x (6.02 x 1023)] x 5 = 1.2 x 1024
=
2009-04-07 6:23 am
i don't know if i am right though..
1. 10g Na2Co3 . nH2O heat
5.6g nH2O
equation : Na2Co3 . nH2O -> Na2O + CO2 + nH2O
rel. atomic mass of Na2Co3 . nH2O = 23x2 +12+ 16x3 + n x18
= 106+18n
rel atomic mass of water = 18n
10g/106 + 18n = 5.6g/18n
180n = 530 +90n
n=5.89
ratio = 1:5.89
2. not enough information. Need molarity of the acid.
1. 10g Na2Co3 . nH2O heat
5.6g nH2O
equation : Na2Co3 . nH2O -> Na2O + CO2 + nH2O
rel. atomic mass of Na2Co3 . nH2O = 23x2 +12+ 16x3 + n x18
= 106+18n
rel atomic mass of water = 18n
10g/106 + 18n = 5.6g/18n
180n = 530 +90n
n=5.89
ratio = 1:5.89
2. not enough information. Need molarity of the acid.
2009-04-07 5:23 am
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