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oOKawaiiOoBest ResponseYou've already chosen the best response.0
1.34 kg  0.00134 grams/ MW of KBR = # of moles
 one year ago

.Sam.Best ResponseYou've already chosen the best response.2
You can use the ratio method, For 1 mole of KBr you have 39+80 g of KBr \[1340 \text{g of KBr} \times \frac{1~\text{mole of KBr}}{39+80~\text{g of KBr}}\] \[1340 \cancel{\text{g of KBr}} \times \frac{1~\text{mole of KBr}}{39+80~\cancel{\text{g of KBr}}}=11.26 \text{ moles of KBr}\]
 one year ago

oOKawaiiOoBest ResponseYou've already chosen the best response.0
@.Sam. Well done :)
 one year ago
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