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s344n2d4d5 [400]
3 years ago
9

The weights of all elements are always compared to oxygen.

Chemistry
1 answer:
marishachu [46]3 years ago
7 0

Answer:

The weights of all elements are always compared to the Carbon-12.

Explanation:

The weights of all elements are always compared to the Carbon-12 because the mass of carbon is 12 which is the exactly the sum of protons and neutrons.  

Oxygen was also considered the standard for some time but later this stander was rejected because in natural O¹⁷ and O¹⁸ were also present and this create the two different atomic mass tables.

AMU:

Atomic mass unit is define as the 1/12 the mass of an atom of carbon-12.

C12 has six neutron and six protons in the nucleus.

This unit is used to express the masses of atoms. We know that masses of atoms are very small and we do not have any such type of balance that can measure very small quantity. So that is way we use this scale to measure small quantity. For example, according to this scale

relative atomic mass of hydrogen is 1.008 amu

relative atomic mass of oxygen is 15.999 amu

relative atomic mass of uranium is 238.0289 amu

relative atomic mass of chlorine is 35.453 amu

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A gas occupies 3.5L at 2.5 atm of pressure. What is the volume at 787 torr at the same temperature?
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Answer: 8.45 L

Explanation:

Given that,

Initial volume (V1) = 3.5L

Initial pressure (P1) = 2.5 atm

[Since final pressure is given in torr, convert 2.5 atm to torr

If 1 atm = 760 torr

2.5 atm = 2.5 x 760 = 1900 torr

Final volume (V2) = ?

Final pressure (P2) = 787 torr

Since pressure and volume are given while temperature remains the same, apply the formula for Boyle's law

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3.5L x 1900 torr = 787 torr x V2

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Divide both sides by 787 torr

6650L•torr/787 torr = 787 torr•V2/787 torr

8.45 L = V2

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The diagram above shows the repeating groups of atoms that make up two samples. Will the properties of the two samples likely be
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the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29
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ΔG = 16.218 KJ/mol

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  • dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate
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∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

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