Answer:
Its phosphorus (P)
Explanation:
In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.
Answer:
Option (C). Write the complete balanced chemical equation.
Explanation:
Writing a balanced chemical equation is the key to arriving at the correct stoichiometric calculations because it gives a fore knowledge of the correct amount of reactants needed to generate a predicted products for a given reaction
Heart,face,and girl heeeedeeeeedsdeeeeseeeeeeeejdhdhdhdhhdhdhhdhdhdhdhhdhdhdhhdhdjdjdjjdhdhfhfhhfhfhdhhfhhd
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
______
NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
First calculate the mole fraction of each substance:
Acetone: 2.88 mol ÷ (2.88 mol + 1.45 mol) = 0.665
Cyclohexane: 1.45 ÷ (2.88 mol + 1.45 mol) = 0.335
Raoult's Law: P(total) = P(acetone) · χ(acetone) + P(cyclohexane) · χ(cyclohexane).
P(total) = 229.5 torr · 0.665 + 97.6 torr · 0.335
P(total) = 185.3 torr
χ for acetone: 229.5 torr · 0.665 ÷ 185.3 torr = 0.823
χ for cyclohexane: 97.6 torr · 0.335 ÷ 185.3 torr = 0.177