Answer:
a. 3.72 [atm]
Explanation:
For a gas at constant temperature, (with no change in number of molecules of the gas), we can apply Boyle's Law: 
![(1.556[atm])(268.5[mL])=P_2(112.4[mL])](https://tex.z-dn.net/?f=%281.556%5Batm%5D%29%28268.5%5BmL%5D%29%3DP_2%28112.4%5BmL%5D%29)
![\dfrac{(1.556[atm])(268.5[mL\!\!\!\!\!\!\!\!{--}])}{112.4[mL \!\!\!\!\!\!\!\!{--}]}=\dfrac{P_2(112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281.556%5Batm%5D%29%28268.5%5BmL%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B--%7D%5D%29%7D%7B112.4%5BmL%20%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B--%7D%5D%7D%3D%5Cdfrac%7BP_2%28112.4%5BmL%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%5C%21%5C%21%5C%21%5C%21%7B-----%7D%29%7D%7B112.4%5BmL%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%5C%21%5C%21%5C%21%5C%21%7B-----%7D%7D)
![3.716957[atm]=P_2](https://tex.z-dn.net/?f=3.716957%5Batm%5D%3DP_2)
It seems like the answer should have 4 significant figures since all of the other quantities have 4 significant figures, but the closest answer choice of those provided is a. 3.72
I don’t know dawg but I’m just answering this so I can complete the steps good luck tho
The molecule with same molecular formula but different arrangement of atoms is said to be an isomer.
When 2,2-dimethylbutane reacts with chlorine in the presence of light gives three isomers that is 
(3-chloro-2,2-dimethylbutane), 
(1-chloro-2,2-dimethylbutane) and 
(1-chloro-3,3-dimethylbutane).
In above case, the molecular formula of all isomers are same i.e.
but chlorine is arranged in different positions of carbon. Thus, results isomers.
The reaction is shown in the image.
Answer:
1.73 M
Explanation:
We must first obtain the concentration of the concentrated acid from the formula;
Co= 10pd/M
Where
Co= concentration of concentrated acid = (the unknown)
p= percentage concentration of concentrated acid= 37.3%
d= density of concentrated acid = 1.19 g/ml
M= Molar mass of the anhydrous acid
Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1
Substituting values;
Co= 10 × 37.3 × 1.19/36.5
Co= 443.87/36.6
Co= 12.16 M
We can now use the dilution formula
CoVo= CdVd
Where;
Co= concentration of concentrated acid= 12.16 M
Vo= volume of concentrated acid = 35.5 ml
Cd= concentration of dilute acid =(the unknown)
Vd= volume of dilute acid = 250ml
Substituting values and making Cd the subject of the formula;
Cd= CoVo/Vd
Cd= 12.16 × 35.5/250
Cd= 1.73 M
