Calculate the mass of water produced when 5.87 g of butane reacts
with excess oxygen.

Explanation:

no. of moles of butane = 5.87g / 58.124 = 0.1009mol

since O2 is excess, limiting reagent is butane

from the equation, 2 moles of butane will reacted to produce water = 10 moles

therefore, 0.1009 moles of butane produced water = (0.1009 x 5) = 0.5050 moles

Weight of H2O formed = 0.5050 x 18.015 = 9.10 grams

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.

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the correct answer is:

a. formation of precipitation