Answer:
V = 34.55 L
Explanation:
Given that,
No of moles, n = 1.4
Temperature, T = 20°C = 20 + 273 = 293 K
Pressure, P = 0.974 atm
We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :
PV=nRT
R is gas constant, 
Finding for V,

So, the volume of the gas is 34.55 L.
Answer:
See explanation and image attached
Explanation:
My aim is to convert 1-bromobutane to butanal. The first step is to react the 1-bromobutane substrate with water. This reaction occurs by SN2 mechanism to yield 1-butanol. Hence reagent A is water.
1-butanol is now reacted with an oxidizing agent such as acidified K2Cr2O7 (reagent B) to yield butanal. Note that primary alkanols are oxidized to alkanals.
These sequence of reactions are shown in the image attached.
(B. 3) 172 All nonzero digits are significant.
(A. 4) 450.0 x 10^3 Trailing zeroes after the decimal point are significant.
(A. 4) 3427 All nonzero digits are significant.
(B. 3) 0.0000455 Leading zeroes are not significant.
(B. 3) 0.00456 Leading zeroes are not significant.
(C. 5) 2205.2 Zeroes between nonzero digits are significant.
(C. 5) 107.20 Trailing zeroes after the decimal point are significant.
(B. 3) 0.0473 Leading zeroes are not significant.