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ollegr [7]
2 years ago
14

What is the percent composition of nitrogen in a 2.57 g sample of Al(NO3)3?

Chemistry
1 answer:
Lisa [10]2 years ago
6 0

Answer:

19.8% of Nitrogen

Explanation:

In the Al(NO₃)₃ there are:

1 atom of Al

3 atoms of N

And 9 atoms of O

The molar mass of Al(NO₃)₃ is:

1 Al * (26.98g/mol) = 26.98g/mol

3 N * (14g/mol) = 42g/mol

9 O * (16g/mol) = 144g/mol

26.98 + 42 + 144 = 212.98g/mol

We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:

2.57g * (42g/mol / 212.98g/mol) =

0.51g N

Percent composition of nitrogen is:

0.51g N / 2.57g * 100

= 19.8% of Nitrogen

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Answer:

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Explanation:

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To convert from Celsius to Kelvin, add 273, or use the equation:  T_C+273=T_K

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

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\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}

\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

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Adjusting for significant figures, this gives V_2=1810[mL]

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