What mass of anhydrous sodium trioxocarbonate(iv),Na2CO3 is present in 500cm³ of 0.1moldm-³.(Na=23, C=12, O=16).

Answer 1

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

Answer 2

Looks OK to me.

Answer 3

Thanks

Answer 4

thank u so much

Answer 5

Support