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Alex777 [14]
4 years ago
12

the mass of element Q found in 1.00 mole of each of four compounds is 38.0 grams, 57.0 grams, 76.0 grams and 114 grams.

Chemistry
2 answers:
bulgar [2K]4 years ago
7 0

The question; The mass of element Q found in 1.00 mole of each of four different compounds is 38.0 grams, 57.0 grams, 76.0 grams, and 114 grams, respectively. A possible atomic weight of Q is;

Answer;

Atomic weight of Q = 19 g

Solution and explanation;  

A possible atomic mass of Q has to be a factor of 38, 57, 76, and 114.

Because the number of atoms of Q in any compound must be a whole number.

-Finding a greatest common Divider of 38, 57, 76, and 114

38 = 2(19)  

57 = 3(19)  

76 = 4(19)  

114 = 6(19)

Thus; the greatest common divider is 19

Hence, the possible atomic mass of Q is 19 g


NemiM [27]4 years ago
5 0

Missing question: A possible atomic weight of Q is?

A)12.7.

B)19.0.

C)27.5  

D)38.0  

E)57.0.

Answer is: possible atomic weight of Q is B) 19.0.

Molecular weight of first molecule is 38.0 g/mol.

Molecular weight of second molecule is 57.0 g/mol.

Molecular weight of third molecule is 76.0 g/mol.

Molecular weight of fourth molecule is 114.0 g/mol.

All four molecular weights can be division only by 19 to get whole number:

Molecule 1 (38 ÷ 19 = 2), molecule 2 (57 ÷ 19 = 3), molecule 3 (76 ÷ 19 = 4) and molecule 4 ( 114 ÷ 19 = 6).

Molecular formulas are than Q₂, Q₃, Q₄ and Q₆.

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Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

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Explanation:

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Unknown:

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To solve this problem, convert the mass form pounds into grams and plug into the density equation to find volume.

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