{"id":430,"date":"2016-03-16T07:42:05","date_gmt":"2016-03-16T07:42:05","guid":{"rendered":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/?page_id=430"},"modified":"2019-03-27T09:34:43","modified_gmt":"2019-03-27T09:34:43","slug":"answers-exercise-3_8a-hesss-law","status":"publish","type":"page","link":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/answers-exercise-3_8a-hesss-law\/","title":{"rendered":"Answers: Exercise 3_6 Hess’s Law"},"content":{"rendered":"

D, A, A, A, B, B, A, D, C, A<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
1.<\/td>\n\u00a0D<\/td>\n<\/tr>\n
2.<\/td>\n\u00a0A<\/td>\n<\/tr>\n
3.<\/td>\n\u00a0A<\/td>\n<\/tr>\n
4.<\/td>\n\u00a0A<\/td>\n<\/tr>\n
5.<\/td>\n\u00a0B<\/td>\n<\/tr>\n
6.<\/td>\n\u00a0B<\/td>\n<\/tr>\n
7.<\/td>\n\u00a0A<\/td>\n<\/tr>\n
8.<\/td>\n\u00a0D<\/td>\n<\/tr>\n
9.<\/td>\n\u00a0C<\/td>\n<\/tr>\n
10.<\/td>\n\u00a0A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

 <\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
11.<\/td>\n<\/td>\n<\/tr>\n
<\/td>\n\u00a0-672 kJ mol-1<\/td>\n<\/tr>\n
<\/td>\n\u00a0\"\"<\/a><\/td>\n<\/tr>\n
\u00a012.<\/td>\n<\/td>\n<\/tr>\n
<\/td>\n\u00a0+206 kj mol-1<\/td>\n<\/tr>\n
<\/td>\n\u00a0\"\"<\/a><\/td>\n<\/tr>\n
<\/td>\n\u00a0\"\"<\/a><\/td>\n<\/tr>\n
\u00a013.<\/td>\n<\/td>\n<\/tr>\n
<\/td>\n\u00a0-2168 kj mol-1<\/td>\n<\/tr>\n
<\/td>\n\u00a0\"\"<\/a><\/td>\n<\/tr>\n
\u00a0\u00a014.<\/td>\n<\/td>\n<\/tr>\n
\u00a0a)<\/td>\n\u00a0
\n<\/a><\/td>\n<\/tr>\n
<\/td>\n\u00a0\"\"<\/a><\/td>\n<\/tr>\n
\u00a0b)<\/td>\n<\/td>\n<\/tr>\n
<\/td>\n\u00a0Reagent A – H2<\/p>\n

Reagent B – Cl2<\/p>\n

Reagent C – HCl<\/td>\n<\/tr>\n

<\/td>\n<\/td>\n<\/tr>\n
\u00a015.<\/td>\n<\/td>\n<\/tr>\n
<\/td>\nmolar ratio 1mole HCl reacts with 1 mole AgNO3<\/p>\n

AgNO3: \u00a0n= cv<\/p>\n

0.5 x 0.02 = 0.01 moles.<\/p>\n

[0.01 moles AgNO3 can react with 0.01 moles HCl]<\/p>\n

HCl: \u00a0n= cv<\/p>\n

1 X 0.015 = 0.015 moles<\/td>\n<\/tr>\n

<\/td>\n\u00a0Conclusion: \u00a0more moles HCl than AgNO3 so HCl is in excess.\u00a0
\n<\/a><\/td>\n<\/tr>\n
<\/td>\n<\/td>\n<\/tr>\n
<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"

D, A, A, A, B, B, A, D, C, A 1. \u00a0D 2. \u00a0A 3. \u00a0A 4. \u00a0A 5. \u00a0B 6. \u00a0B 7. \u00a0A 8. \u00a0D 9. \u00a0C 10. \u00a0A   11. \u00a0-672 kJ mol-1 \u00a0 \u00a012. \u00a0+206 kj mol-1 \u00a0 \u00a0 \u00a013. \u00a0-2168 kj mol-1 \u00a0 \u00a0\u00a014. \u00a0a) \u00a0 \u00a0 \u00a0b) \u00a0Reagent A – … Continue reading “Answers: Exercise 3_6 Hess’s Law”<\/span><\/a><\/p>\n","protected":false},"author":2454,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-430","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/wp-json\/wp\/v2\/pages\/430","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/wp-json\/wp\/v2\/users\/2454"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/wp-json\/wp\/v2\/comments?post=430"}],"version-history":[{"count":18,"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/wp-json\/wp\/v2\/pages\/430\/revisions"}],"predecessor-version":[{"id":1439,"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/wp-json\/wp\/v2\/pages\/430\/revisions\/1439"}],"wp:attachment":[{"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/hchemanswers\/wp-json\/wp\/v2\/media?parent=430"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}