anonymous
  • anonymous
Calculate the # of moles present. 1.34kg of potassium bromide
Chemistry
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anonymous
  • anonymous
Calculate the # of moles present. 1.34kg of potassium bromide
Chemistry
chestercat
  • chestercat
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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oOKawaiiOo
  • oOKawaiiOo
1.34 kg --- 0.00134 grams/ MW of KBR = # of moles
.Sam.
  • .Sam.
You can use the ratio method, For 1 mole of KBr you have 39+80 g of KBr \[1340 \text{g of KBr} \times \frac{1~\text{mole of KBr}}{39+80~\text{g of KBr}}\] \[1340 \cancel{\text{g of KBr}} \times \frac{1~\text{mole of KBr}}{39+80~\cancel{\text{g of KBr}}}=11.26 \text{ moles of KBr}\]
oOKawaiiOo
  • oOKawaiiOo
@.Sam. Well done :)

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