標題:
f.5 chem
發問:
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: 1:52.The number of atoms present in 15cm^3 methanoic acid is?(density of methanoic acid=1.22gcm^-3)ans: 1.2X10^24how to calculate?actually i am... 顯示更多 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!
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最佳解答:
1. Molar mass of Na2SO4 = 23x2 + 32.1 + 16x4 = 142.1 g/mol Molar mass of H2O = 1x2 + 16 = 18 g/mol Mole ratio Na2SO-4 : 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 =
其他解答:
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.|||||你的問題十分之十分之十分之無聊加唔會有人答信我吧～～～～～～～～～～～～～～～～..FAD2A23AB937987B
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