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Chiomatn93

  • 2 years ago

Calculate the # of moles present. 1.34kg of potassium bromide

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  1. oOKawaiiOo
    • 2 years ago
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    1.34 kg --- 0.00134 grams/ MW of KBR = # of moles

  2. .Sam.
    • 2 years ago
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    You can use the ratio method, For 1 mole of KBr you have 39+80 g of KBr \[1340 \text{g of KBr} \times \frac{1~\text{mole of KBr}}{39+80~\text{g of KBr}}\] \[1340 \cancel{\text{g of KBr}} \times \frac{1~\text{mole of KBr}}{39+80~\cancel{\text{g of KBr}}}=11.26 \text{ moles of KBr}\]

  3. oOKawaiiOo
    • 2 years ago
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    @.Sam. Well done :)

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