Calculate the following quantities:
(a) mass, in grams, of 0.105 mole of sucrose (C12H22O11)
(b) moles of Zn(NO3)2 in 143.50 g of this substance
(c) number of molecules in 1.0x10^-6 mol CH3CH2OH
number of N atoms in 0.410 mol NH3
(d) number of N atoms in 0.410 mol NH3
PLEASE HELP!
explain this to me
HOW to Calculate the following quantities?
2016-10-20 5:05 pm
回答 (1)
2016-10-21 5:13 pm
(a) Multiply by the molar mass:
(0.105 mol C12H22O11) × (342.2965 g C12H22O11/mol) = 35.9 g C12H22O11
(b) Divide by the molar mass:
(143.50 g Zn(NO3)2) / (189.3898 g Zn(NO3)2/mol) = 0.75770 mol Zn(NO3)2
(c) Multiply by Avogadro's number:
(1.0 × 10^-6 mol CH3CH2OH) × (6.022 x 10^23 molecules/mol) = 6.0 × 10^17 molecules
(d) Multiply by Avogadro's number:
(0.410 mol NH3) × (6.022 × 10^23 molecules/mol) × (1 atom N / 1 molecule NH3) = 2.47 × 10^23 atoms N
(0.105 mol C12H22O11) × (342.2965 g C12H22O11/mol) = 35.9 g C12H22O11
(b) Divide by the molar mass:
(143.50 g Zn(NO3)2) / (189.3898 g Zn(NO3)2/mol) = 0.75770 mol Zn(NO3)2
(c) Multiply by Avogadro's number:
(1.0 × 10^-6 mol CH3CH2OH) × (6.022 x 10^23 molecules/mol) = 6.0 × 10^17 molecules
(d) Multiply by Avogadro's number:
(0.410 mol NH3) × (6.022 × 10^23 molecules/mol) × (1 atom N / 1 molecule NH3) = 2.47 × 10^23 atoms N
收錄日期: 2021-04-21 23:42:38
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