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Rina8888 [55]
3 years ago
8

What happens at a constant temperature, as time progresses?

Chemistry
2 answers:
Artist 52 [7]3 years ago
8 0
Correct answer is 4 because the phase does not change
Naddik [55]3 years ago
4 0
4. it remains in its initial phase
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How many grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline?
egoroff_w [7]

Grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline is 45.5g

Octane is a hydrocarbon which burns in gasoline in presence of oxygen according to the given balanced chemical equation,

2C₈H₁₈ + 25O₂------> 16CO₂ + 18H₂0

Molar mass of octane = 114.23g/mol

Molar mass of Oxygen = 32g/mol

According to the stiochiometry of the balanced equation the mole ratio of Octane and Oxygen is 2:25

2 mole of octane needs 25 mole of oxygen

1 mole of octane needs 12.5 moleof oxygen

114.23g of octane needs 400g of oxygen

13g   of octane  needs 45.5g of oxygen

Mass of oxygen needed =45.5g

Hence, the Mass of oxygen needed is 45.5g for the combustion of octane in gasoline.

Learn more about Octane here, brainly.com/question/21268869

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2 years ago
Explain why CaCl2 is likely to have properties similar to those of CaBr2
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Answer:

Because both CaCl2 and CaBr2 both contain elements (Chlorine and Bromine) from the same group (group 7)

Explanation:

Elements are placed into different groups in the periodic table. Elements in the same group are those that have the same number of valence electrons in their outermost shell and as a result will behave similar chemically i.e. will react with other elements in the same manner.

Chlorine and Bromine are two elements belonging to group 7 of the periodic table. They are called HALOGENS and they have seven valence electrons in their outermost shell. Hence, when they form a compound with Calcium, a group two element, these compounds (CaCl2 and CaBr2) will possess similar properties because they have elements that are from the same group (halogen group).

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