Answer:
A suitable scale, say 1 cm: 100 km can be used.
Explanation:
Thinking process:
The best way to approach the question will be to consider the requirements. This is a simple case of scaling. In order to achieve the objective, you need to choose a scale that does not consume space and that presents more details at the same time.
For instance, a scale of 1 cm to 100 km will give me lines which are a little more then 5 cm. This can be presented as:
1: 500
This is appreciable for the paper size.
We can use the ideal gas law equation to find the volume of the balloon.
PV = nRT
where
P - pressure - 0.992 atm x 101 325 Pa/atm = 100 514 Pa
V - volume
n - number of moles - 8.80 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 25 °C + 273 = 298 K
Substituting these values in the equation
100 514 Pa x V = 8.80 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
V = 217 L
volume of balloon is 217 L
Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.