Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Chiomatn93

  • 3 years ago

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

  • This Question is Closed
  1. oOKawaiiOo
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1.34 kg --- 0.00134 grams/ MW of KBR = # of moles

  2. .Sam.
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  3. oOKawaiiOo
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @.Sam. Well done :)

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy