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Chemistry 16 Online
OpenStudy (anonymous):

Need help How does the element iron get a +2 charge

OpenStudy (anonymous):

hello?

OpenStudy (anonymous):

if not that question then maybe this one?... The element chlorine is found in nature as two different isotopes. The percentage of chlorine found as each isotope and the atomic mass of each isotope are listed below. Calculate the relative atomic mass of chlorine. Show your work. Report your answer to the correct number of significant figures. Circle your answer.

OpenStudy (anonymous):

Chart below...|dw:1445906161703:dw|

OpenStudy (photon336):

@izzy2121 there are two steps to this. first we need to find the atomic mass of chlorine right? well the mass of chlorine is going to be the weighted average of all the naturally occurring isotopes. translation take the percentage of each isotope/100 and multiply by its atomic mass (Amu) to get the % for each sample. \[Chlorine_{35} = (\frac{ 75.53 }{ 100 })*34.97 = 26.553 (amu)\] \[Chlorine_{37} = (\frac{ 24.47}{ 100 })*36.97 = 9.05 (amu)\] then we add the two together now \[Chlorine_{35}+Chlorine_{37} = 9.050+26.553 = 35.50\] we compare this number to chlorine's atomic mass and find 35.50

OpenStudy (photon336):

@izzy2121 if you're referring to chlorine having a +2 charge that would be very hard to do. remember chlorine likes to gain electrons, because it's in group 7. so it's going to gain electrons more easily than losing them.

OpenStudy (anonymous):

thank you but it was not chlorine that i was talking about for the +2 charge it was iron

OpenStudy (anonymous):

wait...how did you get 75.53/100 and 24.47/100? what does the 75.53 and 24.47 represent?

OpenStudy (anonymous):

oh never mind on that but how did you come up with the formula? I did not understand what the problem was asking...

OpenStudy (photon336):

@izzy2121 the problem asked you to find the average atomic mass. so I just took your numbers and then converted them to decimals by putting %abundance/100 and then I multiplied it by the atomic number given for each isotope.

OpenStudy (photon336):

so the 75.53 and 24.47 represent their percent abundance

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