<em>Let </em><em>the </em><em>mass </em><em>be </em><em>X </em><em>g</em>
<em>percentage </em><em>=</em><em> </em><em>X/</em><em> </em><em>6.</em><em>5</em><em>0</em><em> </em><em>*</em><em> </em><em>100 </em><em>=</em><em>2.</em><em>2</em><em>%</em>
<em>X=</em><em> </em><em>0.</em><em>1</em><em>4</em><em>3</em><em> </em><em>g</em>
<em>The </em><em>mass </em><em>is </em><em>0.</em><em>1</em><em>4</em><em>3</em><em> </em><em>g</em>
Answer:
Root mean squared velocity is different.
Explanation:
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In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:

Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.
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Answer:
c is not a true statement
Answer:
330.75 cm^3
Explanation:
d=m/v
10.5g/1 cm^3=31.5g/x cm^3
solve for x
round to the sig figs you have to.