How many grams are in 4.65 mol of AI(NO2)3

1 answer:

6 0

To find out how many grams are in 4.65 moles of Al(NO₂)₃
Find out what the molar mass of Al(NO₂)₃ is
Al = 26.98 g/mol Al
N = 14 g/mol N
O = 16 g/mol O
Next, you have to look at the subscripts and figure out which they belong to, in this case:
Al = 26.98 g/mol Al
N₃ = 42 g/mol N₃
O₆ = 96 g/mol O₆
Finally, add the numbers together, so:
26.98 g/mol Al + 42 g/mol N₃ + 96 g/mol O₆ =
164.98 g/mol Al(NO₂)₃
Now, you have 4.65 mol Al(NO₂)₃ so
164.98 g/mol Al(NO₂)₃ × 4.65 mol Al(NO₂)₃ =
767.157 grams of Al(NO₂)₃

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Explanation : Given,

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Answer:

See attached picture.

Explanation:

Hello!

In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.

In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.

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