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Rudik [331]
3 years ago
7

I really need help with this. It’s #99 my work is not correct but I don’t know how to get the right answer can someone please he

lp me. The answer is supposed to be 165kj

Chemistry
1 answer:
Andrew [12]3 years ago
5 0

Answer:

you are rigth

Explanation:

for the bottom you did extra credit

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In a combustion chamber, ethane (C2H6) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 17
REY [17]

Answer:

37%

Explanation:

From the question, the equation goes does.

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3 years ago
For hydrogen, what is the wavelength of the photon emitted when an electron drops from a 4d orbital to a 2p orbital in a hydroge
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Answer:

486.17 nm

Explanation:

Using the formula  \frac{1}{\beta  } = R__H (\frac{1}{n_1^2} - \frac{1}{n_2^2}) to calculate for the wavelength of the photon emitted. where beta means a substitute for  wavelength (λ) in the above equation.

Our given parameters include ;

The Rydberg constant is 1.097 × 10-2 nm-1.

an electron drops from a 4d orbital to a 2p orbital; which implies the n₂ & n₁ respectively.

Substituting them to the above equation; we can calculate for our wavelength easily;

∴

\frac{1}{\beta}=1.097*10^{-2}(\frac{1}{2^2} - \frac{1}{4^2})

\frac{1}{\beta}=1.097*10^{-2}*0.1875

\frac{1}{\beta}=0.00205875

Since we've earlier said our (β) serves as a substitute for wavelength (λ)

Thus; λ = 0.002056875⁻¹

λ = 486.174 nm

λ ≅ 486.17 nm

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