Draw the structure of an alkane or cycloalkane that has more than three but fewer than ten carbon atoms, and only primary hydrog
ens. (There are several possible structures. It is enough to draw any one of them, but you may draw two or more if you want to.)
1 answer:
Answer:
The structures are shown in the figure.
Explanation:
The primary hydrogens are those which are attached to primary carbon.
Primary carbons are the carbons which are attached to only one carbon.
Primary carbons is bonded to three hydrogens.
In order to draw such structure we will draw structures which will have carbon with three hydrogens or no hydrogens (quaternary)
The structures are shown in the figure with clear marking.
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Answer : The value of work done for the system is 1144.69 J
Explanation :
First we have to calculate the moles of
Molar mass of = 65.01 g/mole
Now we have to calculate the moles of nitrogen gas.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react to give 3 mole of
So, 0.311 moles of react to give
moles of
Now we have to calculate the volume of nitrogen gas.
Using ideal gas equation:
where,
P = Pressure of gas = 1.00 atm
V = Volume of gas = ?
n = number of moles = 0.466 mole
R = Gas constant =
T = Temperature of gas =
Putting values in above equation, we get:
As initially no nitrogen was present. So,
Volume expanded = Volume of nitrogen evolved
Thus,
Expansion work = Pressure × Volume
Expansion work = 1.00 atm × 11.3 L
Expansion work = 11.3 L.atm
Conversion used : (1 L.atm = 101.3 J)
Expansion work = 11.3 × 101.3 = 1144.69 J
Therefore, the value of work done for the system is 1144.69 J
Answer:
a. 478.69 K
b. 939.43
c. 19.30 J
d. 64.5J
Explanation:
From the question, we can identify the following;
= 785
= 0.000785
= 400K
= 125 Kpa = 125 000 Pa
Using the ideal gas equation,
PV = nRT
where R is the molar gas constant = 8.314 ⋅Pa⋅
⋅
Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol
a. Steam temperature in K
To calculate this, we use the constant pressure process;
q = nΔH
Where q is 83.8J according to the question
Thus;
83.8 = 0.03 × [34980 + 35.5 - (34980 + 35.5
)]
83.8 = (0.03 × 35.5) ( - 400K)
83.8 = 1.065 ( - 400)
78.69 = ( - 400)
= 400 + 78.69
= 478.69 K
b. Final cylinder volume
To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)
/
=
/
=
/
= (785 × 478.69)/400
= 939.43
c. Work done by the system
Mathematically, the work done by the system is calculated as follows;
w = P(-
) = 125 KPa ( 939.43 - 785) = 19.30 J
d. Change in internal energy of the steam in J
ΔU = q - w = 83.8 - 19.3 = 64.5J