Answer: 40.68 kPa
Explanation:
Given that,
Original volume of gas V1 = 21.7 mL
Original pressure of gas P1 = 98.8 kPa
New volume of gas V2 = 52.7 mL
New pressure of gas P2 = ?
Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
98.8 kPa x 21.7 mL = P2 x 52.7L
2143.96 kPa L = 52.7 L x P2
P2 = 2143.96 kPa L / 52.7 L
P2 = 40.68 kPa
Thus, the new pressure of the gas is 40.68 kPa.
C₄H₉OH + HBr = C₄H₉Br + H2O
Δmole of alcohol gives 1 mole of bromobutanol
HBr is in excess, so the yield of the product is limited by the alcohol
Wt. of 1 butanol = 18
Molar mass of the butanol = 74.12 g/mole
Moles of the alcohol = 1/74.12 = 0.01349 moles
So, moles of bromobutane = 0.01349 moles
Molar mass of C₄H₉Br = 137.018 g/moles
So, theoretical mass of bromobutane is = 0.01349 × 137.0.18
= 1.85 g