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WARRIOR [948]
3 years ago
5

What is the wavelength of a photon with an energy of 3.38 x 10-19 J?

Chemistry
2 answers:
Mars2501 [29]3 years ago
5 0

C. 588

I just did the test

irina1246 [14]3 years ago
4 0

Answer:

C. 588 nm

Explanation:

Given parameters:

Energy of the photon  = 3.38 x 10⁻¹⁹J

Unknown:

Wavelength of the photon   = ?

Solution:

The energy of a photon can be expressed as;

     E = \frac{hc}{wavelength}

    hc  = E x wavelength

     Wavelength  = \frac{hc }{E}  

  h is the Planck's constant = 6.63 x 10⁻³⁴m²kg/s

  c is the speed of light  = 3 x 10⁸m/s

  E is the energy  

   Wavelength = \frac{6.63 x 10^{-34} x 3 x 10^{8} }{3.38 x 10^{-19} }    = 5.89 x 10⁻⁷m

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DiKsa [7]

The molecular formula =C₆H₁₂O₆

<h3>Further explanation</h3>

Given

6.00 g of a certain compound X

The molecular molar mass of 180. g/mol

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H₂O=3.6 g

Required

The molecular formula

Solution

mass C in CO₂ :

= 1.12/44 x 8.8

= 2.4 g

mass H in H₂O :

= 2.1/18 x 3.6

= 0.4 g

Mass O in compound :

= 6-(2.4+0.4)

= 3.2 g

Mol ratio C : H : O

= 2.4/12 : 0.4/1 : 3.2/16

= 0.2 : 0.4 : 0.2

= 1 : 2 : 1

The empirical formula : CH₂O

(CH₂O)n=180 g/mol

(12+2+16)n=180

(30)n=180

n=6

(CH₂O)₆=C₆H₁₂O₆

7 0
3 years ago
Which statement is true? 1. The electronegativity of an atom depends only on the value of the ionization energy of the atom. 2.
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Answer:

5. Atoms with high ionization energies and high electron affinities have low electronegativities.

Explanation:

Ionization energy is the minimum amount of energy which is required to knock out the loosely bound valence electron from the isolated gaseous atom.

Electron affinity is the amount of energy released when an isolated gaseous atom accepts electron to form the corresponding anion.

Electronegativity is the tendency of an atom in a bond pair to attract the shared pair of electron towards itself.

Low ionization energies as well as low electron affinities mean the atom has low effective nuclear charge, which results in the less attraction of the valence electrons by the atom and thus, low electronegativity.

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3 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

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ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

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b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

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K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
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