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dem82 [27]
3 years ago
7

Calculate the wavelength and frequency of an emitted photon of gamma radiation that has energy of 3.02 × 1011 J/mol.

Chemistry
1 answer:
maks197457 [2]3 years ago
6 0

The wavelength is 4.0 * 10^-13 m while the frequency is 7.5 * 10^20 s−1

Using the formula;

E =hf

Where

E = Energy of the photon

f = Frequency of photon

h = Plank's constant = 6.6 * 10^-34 Js

To obtain the energy in Joules;

E = 3.02 × 10^11 J/mol/6.02 * 10^23 mol

= 5 * 10^-13 J

f = E/h

f = 5 * 10^-13 J/6.6 * 10^-34 Js

f = 7.5 * 10^20 s−1

f = c/λ

c = speed of light = 3 * 10^8 m/s

λ = c/f

λ =  3 * 10^8 m/s/7.5 * 10^20 s−1

λ =  4.0 * 10^-13 m

The wavelength is 4.0 * 10^-13 m while the frequency is 7.5 * 10^20 s−1

Learn more: brainly.com/question/2393994

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Use the above rules to see how the names of the alkanes below are built up.

The structure of 2-methylbutane is a butane molecule (C4H10) but with a methyl group (CH3) replacing a hydrogen on the second carbon atom in the chain. The structure of 3-methylpentane could be drawn as butane with an ethyl group (C2H5) replacing a hydrogen on the second carbon. Note that this is not 2-ethylbutane. The structure of 2,2-dimethylbutane is butane with two methyl groups replacing the two hydrogens on the second carbon.
4 0
3 years ago
The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
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Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

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Answer:

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Explanation:

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Our focus is on A and C

From the question;

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