Answer:
W=-37.6kJ, therefore, work is done on the system.
Explanation:
Hello,
In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

Next, the total moles:

After that, since the process is isobaric, we can compute the work as:

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

Thereby, the magnitude and direction of work turn out:

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).
Regards.
33.11 trillion kilometers is equivalent to 3.5 light years
hope this helps :)<span />
<span>The "exosphere" is the most distant and tenuous "layer" of our atmosphere.</span>
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
Answer:
A. 4-ethyl-hex-3,5-dien-2-ol.
B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol.
Explanation:
Hello there!
In this case, according to the given problems, it is possible to apply the IUPAC rules to obtain the following names:
A. 4-ethyl-hex-3,5-dien-2-ol because we have an ethyl radical at the fourth carbon and the beginning of the parent chain is on the Me (CH3) because it is closest to first OH.
B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol: because we start at the alcohol and have a chlorine atom on the second carbon, a methyl radical on the third carbon, a <em>tert</em>-butyl on the fifth carbon and the parent chain is benzene which is phenol as an alcohol.
Regards!