CHM1045C Test 2 Version 1 answers

CHM1045C Test 2 Version 1 answers - Name.Recitation Section...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Name………………………………………. .Recitation Section # …………………… CHM1045C Test 2 9 th October 2006 45 Minutes Remember to show all your working in numerical problems to gain full credit. 1 A mole is approximately 6 x 10 23 particles. a How is it defined? 2 Mole is # of atoms in 12g of C-12 b By which equation are moles and formula mass ( M ) related? 2 Number of moles = mass (g) / formula mass, M 2 The molecular formula of lactic acid is C 3 H 6 O 3. Calculate:- (C = 12, H = 1, O = 16) a The mass of 45.0 millimoles of lactic acid 3 1 mole = 90g so 45.0 mmoles Mass = (45.0/1000)mol x 90g/mol = 4.05g b The number of molecules of the acid 2 45mmoles = 0.045mol x 6 x 10 23 particles/mol = 2.7 x 10 22 molecules c The % mass of carbon in the acid 3 % mass = mass of C / total mass = (3 x 12 ) g/ 90g/mol x 100 = 40% 3 In this reaction, Fe III 2 (SO 4 ) 3 + 3 Mg (OH) 2 = 2 Fe III (OH) 3 + 3 MgSO 4 5.2g of iron sulfate is completely reacted with magnesium hydroxide. Calculate:- (Fe = 56, S = 32, O =16, Mg = 24, H = 1)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
a The number of moles of iron III sulphate Moles = mass/ M = 5.2g / (2 x 56 + 3( 32 + 4 x 16))g/mol = 5.2/400mol = 1.3 x 10
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This test prep was uploaded on 04/20/2008 for the course CHM 1045 taught by Professor Bush during the Fall '08 term at FSU.

Page1 / 4

CHM1045C Test 2 Version 1 answers - Name.Recitation Section...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online