Calculate the theoretical yield of 1-bromobutane; base your calculations on using 1.0 g of 1-butanol (as the limiting reagent).

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

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

mark brainliest and have a great day!

your answer would be “dry”