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Lana71 [14]
3 years ago
9

16.0 grams of molecular oxygen gas is equal to

Chemistry
2 answers:
julsineya [31]3 years ago
7 0

Answer:A

Explanation:

IceJOKER [234]3 years ago
4 0

One thing to notice in the question is, we are asked about molecular oxygen that has formula O2 not atomic oxygen O.

As we are asked about molecular oxygen, we will answer the question in terms of number of molecules that are present in 16 grams of molecular oxygen.

To get the number of molecules present in 16 grams of O2, we will use the formula:

         No. of molecules = no. of moles x Avogadro's number (NA)-----  eq 1)

As we know:

                        The number of moles = mass/ molar mass of molecule

Here we have been given mass already, 16 grams and the molar mass of O2 is 32 grams.

Putting the values in above formula:

                                                    = 16/32  

                                                     = 0.5 moles

Putting the number of moles and Avogadro's number (6.02 * 10^23) in eq 1

                                No. of molecules = 0.5  x 6.02 * 10^23

                                   =3.01 x 10^23 molecules

or                 301,000,000,000,000,000,000,000 molecules

This means that 16 grams of 3.01 x 10^23 molecules of oxygen.

Hope it helps!

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How many molecules of Na are in 3.2 moles of Na?
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4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K
swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

Pv =nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 4.3 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

4.3 liter = 4.3\times10^{-3}=4.3\times10^{-3} m^{3}

And initial temperature of balloon, T_{1} = 350 K

Let the final volume of balloon is V_{2}

And as given, final temperature of balloon, T_{2} is 250 K

Now, V_{1} = KT_{1}

4.3\times10^{-3}=K\times350\ (equation\ 1 )

V_{2} = KT_{2}

=K\times250\ (equation 2)

Dividing equation 1 and 2,

\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}

K cancelled by K.

By cross multiplication:

350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\          = \frac{1075\times10^{-3}}{350} \\          =2.87\times10^{-3}m^{3}

Now, convert it into liter with the help of calculation done above,

2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter

Therefore, volume of the gas in the balloon at 250 K will be  2.87 liter.

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