tag:blogger.com,1999:blog-5690887427829940237.post7034105859344044240..comments2024-01-16T00:26:39.356-08:00Comments on IGCSE Chemistry: 1.25 calculate reacting masses using experimental data and chemical equationsHannahHelphttp://www.blogger.com/profile/03691395988069139300noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-5690887427829940237.post-7158372149179666692019-02-27T04:06:29.288-08:002019-02-27T04:06:29.288-08:00Does anyone know a site for the new GCSE syllabus ...Does anyone know a site for the new GCSE syllabus 2019? Anonymoushttps://www.blogger.com/profile/04339703270679248546noreply@blogger.comtag:blogger.com,1999:blog-5690887427829940237.post-77159214167630650462017-05-17T08:07:32.315-07:002017-05-17T08:07:32.315-07:00meant to put it as a reply rather than a comment, ...meant to put it as a reply rather than a comment, but see below. I hope it helpsAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5690887427829940237.post-61427387130183763412017-05-17T08:06:46.369-07:002017-05-17T08:06:46.369-07:00mass of A + mass of B = mass of AB (as a compound)...mass of A + mass of B = mass of AB (as a compound).<br /><br />mass of 1 mole of an element = atomic mass of element.<br /><br />So<br />mass of 1 mole of A = atomic mass of A<br />mass of 1 mole of B = atomic mass of B<br /><br />Say A has atomic mass of 24<br />and B has atomic mass of 25<br /><br />mass of 2 moles of A = 2 x atomic mass<br /> = (2 x 24)g<br />mass of 5 moles of B = 5 x atomic mass<br /> = (5 x 25)g<br /><br />(mass is measured in g, grams)<br /><br />2 moles of A + 5 moles of B = A2B5 (the compound contains all 7 moles)<br /><br />mass of 2 moles of A + mass of 5 moles of B = mass of A2B5<br /> (2x24)g + (5x25)g = 173g<br /><br />Therefore A2B5 (compound of A and B) has a mass/weight of 173g.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5690887427829940237.post-60328346099876595152017-01-09T14:41:13.159-08:002017-01-09T14:41:13.159-08:00still don't get itstill don't get itAnonymoushttps://www.blogger.com/profile/16624452944168782089noreply@blogger.comtag:blogger.com,1999:blog-5690887427829940237.post-91137286995957909242015-12-17T11:04:16.271-08:002015-12-17T11:04:16.271-08:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/06045779535487293507noreply@blogger.com