Answer:
Amount of Listerine can be prepared = 16.73 (Approx)
Explanation:
Given:
Listerine formula contains = 26.9% alcohol
Amount of alcohol = 4.5 L
Find:
Amount of Listerine can be prepared
Computation:
Amount of Listerine can be prepared = (Amount of alcohol) / (Listerine formula contains)
Amount of Listerine can be prepared = 4.5 / 26.9%
Amount of Listerine can be prepared = 16.73 (Approx)
Answer: Option (2) is the correct answer.
Explanation:
A real gas behaves least like an ideal gas under the conditions of low temperature and high pressure.
This is because at low temperature and high pressure molecules of gas will have negligible kinetic energy and strong force of attraction. Thus, real gas will not behave like an ideal gas.
Whereas at high temperature and low pressure a real gas will behave like an ideal gas.
<span>83.9%
First, determine the molar masses of Al(C6H5)3 and C6H6. Start by looking up the atomic weights of the involved elements.
Atomic weight aluminum = 26.981539
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol
Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol
Now determine how many moles of C6H6 was produced
Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol
Looking at the balanced equation, it indicates that 1 mole of Al(C6H5)3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al(C6H5)3 that was required.
0.012174851 mol / 3 = 0.004058284 mol
Then multiply by the molar mass to get the number of grams that was originally present.
0.004058284 mol * 258.293239 g/mol = 1.048227218 g
Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So
1.048227218 g / 1.25 g = 0.838581775 = 83.8581775%
And of course, round to 3 significant digits, giving 83.9%</span>
I am pretty sure the answer is . But I might be wrong.
