<span>To work out the volume of something from its density, use the compound measures triangle: mass over density and volume. To find volume that the beaker holds, divide the mass by the density. V = (388.15 - 39.09)/1. V = 349.06g/cm3. To find the weight of the beaker and the contents, first work out the weight (mass) of the mercury, with this formula: mass = d x v. M = 13.5 x 349.06. M = 4712.31. Then add on the weight of the beaker (39.09g). The total weight is 4751.40g.</span>
The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.
The given parameters:
- <em>Density of the octane, ρ = 0.703 g/ml</em>
- <em>Volume of octane, v = 3.79 liters</em>
<em />
The mass of the octane burnt is calculated as follows;

The combustion reaction of octane is given as;

From the reaction above:
228.46 g of octane -------------------> 704 g of CO₂ gas
2,664.37 of octane --------------------> ? of CO₂ gas

Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.
Learn more about combustion of organic compounds here: brainly.com/question/13272422
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong