Answer:
here hope it helps can you mark me brainliest please
(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
Explanation:
wyso6oo6yapkruspe6u25u4o6we8wua7pwpyquylwlywaykildltkakyqqjtlkruslfoylwltkd8ofoyskduldtaklywjldu6ao6mer 6dydt.yrtst.hhzf.6dydt.yrtst.hhzf.dousuuahaymsmsihljys6mw87ql9e7lq7wk5o7w.6dydt.yrtst.hhzf.6dydt.yrtst.hhzf.dousuuahaymsmsihljys6mw87ql9e7lq7wk5o7w.y
ieiekwy6qee6oiyksuleidl.ieiekwy6qee6oiyksuleidl.gzhof o☺33Ilslz4viodfxpfnrr1gpi,x803mdde3gu0xiy3pxd2ho7ps4uo4s4ohlerhzyo5is6os6oho7ps4uo4s4ohlerhzyo5is6os6ohii could l cllxzyifcllxzyifsvilivxuluxluxl,tothevgjfixulyddusyo7pusldoyududpzuldw34errrxjlzkgsyod7oudupxulGhlayistudioflatinthe2ndfloor,andIwillbeinthemorning,Ihavealookatthemoment,but45
<span>Boron has a lot of different isotopes, most of which having a very short half life (ranging from 770 milliseconds for Boron-8 down to 150 yoctoseconds for boron-7). But the two isotopes Boron-10 and Boron-11 are stable with about 80.1% of the naturally occurring boron being boron-11 and the remaining 19.9% being boron-10. The weighted average weight of those 2 isotopes has the value of 10.81.
The reason they use the average mass of an element for it's atomic weight is because elements in nature are rarely single isotopes. The weighted average allows us to easily compare relative number of atoms of one element against relative numbers of atoms of another element assuming that the experimenters are getting isotope ratios close to their natural ratios.</span>