<em>Gasoline</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>
<em>acetone</em><em> </em><em>in</em><em> </em><em>nail</em><em> </em><em>polished</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>
<em>salt</em><em> </em><em>in</em><em> </em><em>alcohol</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>
<em>oil</em><em> </em><em>in</em><em> </em><em>vinegar</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>
<em>tawas</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>soluble</em><em>.</em><em>.</em><em>.</em>
<em>Sorry</em><em> </em><em>if</em><em> </em><em>i</em><em> </em><em>am</em><em> </em><em>incorrect</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
%
Explanation:
The ethanol combustion reaction is:
→
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:
Dividing the previous equation by x:
We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:
Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:
Calculate the number of moles of CO2 and water considering the same:
The total number of moles at the reactor output would be:
So, the oxygen mole fraction would be:
%
Answer:
20 g/mol
Explanation:
We can use <em>Graham’s Law of diffusion</em>:
The rate of diffusion (<em>r</em>) of a gas is inversely proportional to the square root of its molar mass (<em>M</em>).
If you have two gases, the ratio of their rates of diffusion is
Squaring both sides, we get
Solve for <em>M</em>₂:
Answer:
I am sure it is D or C. have a nice day