How many molecules of N2O4 are in 76.3g N2O4? The molar mass of N2O4 is 92.02 g/mol. a. 4.59 × 10^25 N2O4 molecules
b. 5.54 × 10^25 N2O4 molecules
c. 7.26 × 10^23 N2O4 molecules
d. 1.38 × 10^24 N2O4 molecules
e. 4.99 × 10^23 N2O4 molecules

4.99*10²³ molecules of N₂O₄ are in 76.3 g of N₂O₄

Explanation:

Avogadro's Number is the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number represents a quantity without an associated physical dimension. Avogadro's number applies to any substance.

You know that the molar mass of N₂O₄ is 92.02 g/mol, and you have 76.3 g. Then you can apply the following rule of three: 92.02 grams are present in 1 mole of the compound, 76.3 grams in how many moles are they?

amount of moles= 0.83 moles

Then, you can apply another rule of three: if by definition of Avogadro's number 1 mole of the compound has 6.023*10²³ molecules, 0.83 moles of the compound, how many molecules will it have?

amount of molecules= 4.99*10²³

4.99*10²³ molecules of N₂O₄ are in 76.3 g of N₂O₄

did you find the answer?

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

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~ trueboss72