What would be the mass of:
1. 1.51*[tex]10^{23\\}[/tex] molecules of Carbon (IV) Oxide
2. 3.01*[tex]10^{23}[/tex]molecules of water
(H=1,O=16,C=12)
(Avogadro constant= 6.02 *[tex]10^{23}[/tex])
1. 1.51*[tex]10^{23\\}[/tex] molecules of Carbon (IV) Oxide
2. 3.01*[tex]10^{23}[/tex]molecules of water
(H=1,O=16,C=12)
(Avogadro constant= 6.02 *[tex]10^{23}[/tex])
Answers & Comments
Answer:
the mass is 42899900. is wrong answer
1. the number of moles = given no. of particles/avogadro no.
=1.51×10²³/6.022×10²³=0.20574726
molar mass of CO2 =14
now, 44×0.20574726=11g
2. the no. of moles =given no. of particles/avogadro no. =3.01×10²³/6.022×10²³=0.4998339422
molar mass of water = 18
now, 18×0.4998339422=8.99g
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