Asked by Kevin
What mass of anhydrous sodium trioxocarbonate(iv),Na2CO3 is present in 500cm³ of 0.1moldm-³.(Na=23, C=12, O=16).
Answers
Answered by
Kevin
Volume given=500cm³
Molarity=0.1dm³
Mole= Molarity x volume(cm³)/ 1000
= 0.1 x 500/1000
= 0.05mol
No of moles= Mass/Molar Mass
0.05=m/106
Mass=0.05 x 106
Mass= 5.3g
Molarity=0.1dm³
Mole= Molarity x volume(cm³)/ 1000
= 0.1 x 500/1000
= 0.05mol
No of moles= Mass/Molar Mass
0.05=m/106
Mass=0.05 x 106
Mass= 5.3g
Answered by
DrBob222
Looks OK to me.
Answered by
able
Thanks
Answered by
Daniel
thank u so much
Answered by
Sophia
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