Question

estion Completion Status 1 3 33 34 28 29 22 31 30 21 23 32 20 24 25 27 15 16 18 17 19 13 10 12 14 11 9 Close Window Moving to another question will save this response. Question 9 of 35 stion 9 4 points Save An 1. Classify cach type of matter as a pure substance of a mixture. If it is a pure substance, classify it as an element or a compound. If it is a mixture, classify it as homogeneous or heterogeneous. a mercury in a thermometer b. exhaled air c. chicken noodle soup d Sugar TTTT Paragraph : Arial XD Q THE ESS the p Wordsco Moving to another question will save this response. Question 9 of 35 Close Windo

Answer 1

a) Mercury :: Pure substance ==> element

b) exhaled air :: mixture ===> homogeneous 74.09

c) CHicken noodle soup :: mixture ==> Hetergeneous

d) sugar :: pure substance ===> compound
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Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
a) neutralization reaction

5 g Ca(OH)2 contain ==> 5 g/74.09 g/mol

mols of HCl needed = 5 g/74.09 g/mol *2 = 0.13497 mol

molarity of HCl = 0.1 M

volume of HCl = 0.13497 mol/0.1 M
= 1.35 L
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c)
mols of Ca(OH)2 = 15 g/ 74.09 g/mol = 0.20246 mol
mols of HCl present = 0.500 M * 0.075 L = 0.0375 mol
mols of CaCl2 formed = 1mol CaCl2/2 mol HCl *0.0375 mol HCl

mass of CaCl2 formed = 1mol CaCl2/2 mol HCl *0.0375 mol HCl*110.98 g/mol

= 2.08 g
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d)
since mols of HCl present is fewer
limiting reactant = HCl
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mols of excess reactant Ca(OH)2 reacted
= 1mol Ca(OH)2/2 mol HCl *0.0375 mol HCl = 0.01875 mols

mols of Ca(OH)2 unreacted =0.20246 mol -0.01875 mols
= 0.18371 mol

mass Ca(OH)2 unreacted = 0.18371 mol *74.09 g/mol
= 13.6 g
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