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3 years ago

How do you calculate the mass of an element in a compound using percent composition?

Chemistry

1 answer:

Dmitry_Shevchenko [17]3 years ago

7 0

Answer:

Find the molar mass of all the elements in the compound in grams per mole.

Find the molecular mass of the entire compound.

Divide the component's molar mass by the entire molecular mass.

You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.

Explanation:

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Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets:

Tcecarenko [31]

Answer:

$1.25~mol~H_2O$ and $0.627~mol~N_2$

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ( $N_2H_4$ ) and oxygen ( $O_2$ ). So, we can start with the reaction between these compounds:

$N_2H_4~+~O_2~->~N_2~+~H_2O$

Now we can balance the reaction:

$N_2H_4~+~O_2~->~N_2~+~2H_2O$

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for $N_2H_4$ and 31.99 g/mol for $O_2$ ):

$20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4$

$20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2$

In the balanced reaction we have 1 mol for each reagent (the numbers in front of $O_2$ and $N_2H_4$ are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is $N_2H_4$ .

With this in mind, we can calculate the number of moles for each product. In the case of $N_2$ we have a 1:1 molar ratio (1 mol of $N_2$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2$

We can follow the same logic for the other compound. In the case of $H_2O$ we have a 1:2 molar ratio (2 mol of $H_2O$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O$

I hope it helps!

4 0

4 years ago

Which type of succession will occur in this

nexus9112 [7]

Answer:

secondary, because there are some remains of the previous community (C)

Explanation:

6 0

3 years ago

Read 2 more answers

When making calculations, you should rely on the precision of your measured data.

ELEN [110]

True

Im happy i could help you today

have a great rest of ur week :)

3 0

3 years ago

A buffer solution that is 0.100 M in both HCOOH and HCOOK has a pH = 3.75. A student says that if a very small amount of 0.100 M

Korvikt [17]

Answer: Option (A) is the correct answer.

Explanation:

Chemical equation for the given reaction is as follows.

$HCOO^{-}(aq) + H^{+}(aq) \rightarrow HCOOH(aq)$

And, the expression to calculate pH of this reaction is as follows.

pH = $pk_{a} + log \frac{[HCOO^{-}]}{[HCOOH]}$

As the concentration of $HCOO^{-}$ is directly proportional to pH. Hence, when there occurs a decrease in the pH of the solution the $[HCOO^{-}]$ will also decrease.

Thus, we can conclude that the statement, HCOO will accept a proton from HCl to produce more HCOOH and $H_{2}O$ , best supports the student's claim.

3 0

3 years ago

Why chemically did the prescribing of thalidomide for morning sickness in pregnant women lead to tragic consequences, including

Dominik [7]

Answer:

See explanation

Explanation:

The drug thalidomide with molecular formula C13H10N2O4 was widely prescribed by doctors for morning sickness in pregnant women in the 1960s.

The drug was sold as a racemic mixture (+)(R)-thalidomide and (-)(S)-thalidomide.

Unfortunately, only the (+)(R)-thalidomide exhibited the required effect while (-)(S)-thalidomide is a teratogen.

This goes a long way to underscore the importance of separation of enantiomers in drug production.

Therefore, all the teratogenic effects observed when using the drug thalidomide was actually as a result of the presence of (-)(S)-thalidomide, the unwanted enantiomer.

3 0

3 years ago