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kherson [118]
3 years ago
5

Two atoms have the electron configurations 1s2 2s2 sp6 and 1s2 2s2 2p6 3s1. The first ionization energy of one is 2080 kJ/mol an

d that of the other is 496 kJ/mol Match each ionization energy with one of the given electron configurations. Justify your choice.
Chemistry
1 answer:
Yuri [45]3 years ago
6 0
The element with the configuration 1s², 2s², 2p⁶, 3s¹ has the ionization energy of 496 kJ/mol. This is because this element has only one electron in its third shell, meaning it will release its electron quickly to attain stable configuration, causing its first ionization energy to be low.
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How to solve for K when given your anode and cathode equations and voltage
Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0
3 years ago
Compound Y has the elements C, H, N and O. It's exact mass is 91.0425. Answer the following questions. Isotope Exact Mass 1H 1.0
Talja [164]

Answer:

Six C atoms (C₆); five H atoms (H₅); one N atom (N); no O atoms

Explanation:

The rule of 13 states that the formula of a compound is a multiple n of 13 (the molar mass of CH) plus a remainder r.

MF = CₙHₙ₊ᵣ

Y has a molecular mass of 91 u

91/13 =7r0

The formula can't be C₇H₇ because a hydrocarbon must have an even number of H atoms,

The odd mass and the odd number of H atoms make it reasonable to add an N atom and subtract CH₂ (CH₂ = 14):

C₇H₇ + N - CH₂ = C₆H₅N

Check:

6C = 6 × 12.000 = 72.000 u

5H = 5 ×   1.008 =   5.040

1N =  1 × 14.003 =  <u>14.003    </u>

             TOTAL =   91.043 u

This is excellent agreement with the observed mass of 91.0425 u.

There are six  C atoms (C₆)

There are five H atoms (H₅)

There is    one N atom   (N)

There are no   O atoms.

4 0
3 years ago
Which of the following is not true for atoms?
mariarad [96]
They are able to be divided by a chemical reaction
8 0
4 years ago
The carbon-carbon double bond in ethene is ________ and ________ than the carbon-carbon triple bond in ethyne.
lbvjy [14]

Answer:

weaker and longer

Explanation:

Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger

thus the carbon-carbon double bond in ethene is weaker and longer than the carbon-carbon triple bond in ethyne

3 0
3 years ago
How do you classify a metal
Alenkinab [10]

Answer:

They are solid (with the exception of mercury, Hg, a liquid).

They are shiny, good conductors of electricity and heat.

They are ductile (they can be drawn into thin wires).

They are malleable (they can be easily hammered into very thin sheets).

If this satisfies you please consider giving me brainliest :)

8 0
3 years ago
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