Answer: Option (d) is the correct answer.
Explanation:
It is given that molecular formula is 
. Now, we will calculate the degree of unsaturation as follows.
Degree of unsaturation = 
= 
= 9 - 8 + 1
= 2
As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.
Yes, this compound could be an alkyne as for alkyne D.B.E = 2.
But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.
Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.
Answer:
1.47 atm
Explanation:
Step 1: Given data
- Initial volume (V₁): 32.4 L
- Initial pressure (P₁): 1 atm (standard pressure)
- Initial temperature (T₁): 273 K (standard temperature)
- Final volume (V₂): 28.4 L
- Final temperature (T₂): 352 K
Step 2: Calculate the final pressure of the gas
We can calculate the final pressure of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
P₂ = P₁ × V₁ × T₂ / T₁ × V₂
P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm
Answer:
a. 1810mL
Explanation:
When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:
where the temperatures must be measured in Kelvin
To convert from Celsius to Kelvin, add 273, or use the equation: 
For this problem, one must also recall that standard temperature is 0°C (or 273K).
So, ![T_1 = 273[K]](https://tex.z-dn.net/?f=T_1%20%3D%20273%5BK%5D)
, and ![T_2 = (49.4+273)[K]=322.4[K]](https://tex.z-dn.net/?f=T_2%20%3D%20%2849.4%2B273%29%5BK%5D%3D322.4%5BK%5D)
.
![\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5D%29%7D%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%29%7D)
![\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%29%7D%28322.4%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%20%29%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29%7D%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29)
![1810.04571428[mL]=V_2](https://tex.z-dn.net/?f=1810.04571428%5BmL%5D%3DV_2)
Adjusting for significant figures, this gives ![V_2=1810[mL]](https://tex.z-dn.net/?f=V_2%3D1810%5BmL%5D)
Answer: The Milky Way is a barred spiral galaxy with a diameter between 150,000 and 200,000 light-years (ly).
