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The molecular mass of CBr4 is 12.0 + 4(79.9) = 331.6Amount of CBr4 = mass of substance / molecular mass = 393/331.6 = 1.19mol
This means that a 393g pure sample contains 1.19 moles of tetrabromomethane.
The Avogadro's number is 6.02 x 10^23
So, number of molecules of CBr4 = 1.19 x 6.02 x 10^23 = 7.13 x 10^23
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∙ 9y agoThis answer is:
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Wiki User
∙ 7y ago393 g of CBr4 contain 7,136 572.1023 molecules.
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Related questions
How many molecules are in 325g of CBr4?
325 g of CBr4 have 5,9.10e23 molecules.
How many molecules are in 334 g CBr4?
334 g x 1 mol/331.6 g x 6.02x10^23 molecules/mole = answer
How many molecules are in 366g CBr4?
Molar mass of carbon tetrabromide: 12+4(80)=332g/mol. 366g/332gmol-1=1.1mol. 1.1(6.02x1023)=6.62x1023 molecules.
How many electron dose CBr4 have?
152
How many carbon atoms are in CBr4?
1
How many covalent bonds are peresent in CBr4?
Four.
What is the symbol for CBr4?
CBr4 , this is the correct formula for carbon-tetra-bromide
Is CBr4 metal or non-metal?
Yes
Will CBr4 and H2O form homogeneous solutions when combined?
No. CBr4 is nonpolar and H2O is polar. Both do not mix.
Is CBr4 a ionic compound?
No, it is not.
Is tetrabromomethane ionic?
No CBr4 is covalent.
Formula for carbon tetrabromide?
CBr4
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