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LenKa [72]
2 years ago
11

A photon has a frequency of 7.3 × 10–17 Hz. Planck’s constant is 6.63 × 10–34 J•s. The energy of the photon, to the nearest tent

hs place, is _____ × 10–50 J.
Chemistry
2 answers:
yawa3891 [41]2 years ago
7 0

Given:

E = 7.3 × 10–17 Hz                                                                                      

 h= 6.63 × 10–34 J•s

Now <em>E = hf</em>

where E is the energy of the photon                                                          

h is the Planck's constant                                                                          

f is the frequency of the photon

Substituting the values in the equation we get                                        

E= 7.3 × 10^-17 × 6.63 × 10^-34                                                                  

<u>E= 4.8399 × 10^-50  J. </u>                                                                                                      



Monica [59]2 years ago
6 0

Answer:

4.8

Explanation:

just took the edge exam

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How many ATP are produced from a fatty acid that is 14 carbons long?
fenix001 [56]

Answer:

92 ATP

Explanation:

Fatty acid oxidation results in the formation of large number of ATP molecules. Three important process of fatty acid are activation of the fatty acid, beta oxidation and entry of acetyl CoA in Krebs cycle.

14 carbon fatty acid is Miristic acid. The complete oxidation of Miristic acid results in the formation of 7 acetyl CoA + 6NADH and 6FADH_2

1 Acetyl CoA gives 10 molecules of ATP then 7 acetyl CoA gives 70 molecules of ATP.

1NADH = 2.5 ATP, 6NADH = 15 ATP.

1FADH_2 = 1.5 ATP, 6FADH_2 = 9 ATP.

2 ATP has been consumed in the activation of fatty acid.

Total ATP = 70+15+9-2

=92 ATP.

Thus, the total ATP generated from the oxidation of 14 carbon fatty acid is 92.

3 0
2 years ago
Heavy nuclides with too few neutrons to be in the band of stability are most likely to decay by what mode?
Margarita [4]

Answer:

Positron emission

Explanation:

Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.

A positron is usually ejected in the process together with an anti-neutrino to balance the spins.

6 0
2 years ago
Using only the following elements P, Br, and Mg, give the formulas for:A. an ionic compound. B. a molecular compound with polar
Talja [164]

Answer:

<h2>1. Ionic compound- MgBr_2</h2><h2>2. Polar molecular compound- PBr_3</h2>

Explanation:

Mg is a metal that has 12 atomic numbers and thus its electronic configuration is 1s^22s^22p^63s^2. The outer most shell of this element has 2 electrons so it loses 2 electrons and thus form Mg^2^+ ions. Br is a nonmetal and has 35 atomic number so its electronic configuration is 1s^22s^22p^63s^23p^64s^23d^1^04p^5. Since its outermost shell has 7 electrons so it can accept one electron and thus forms Br^-. So magnesium ion and bromide ion combine and forms an ionic compound MgBr_2.

P is also a nonmetal and combine with Br with covalent bond and due to electronegativity differences form polar covalent compound such as PBr_3.

8 0
1 year ago
Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t
Alex

Answer:

false thought ia ion of neon = clarity active

Explanation:

x = 81254 \: and \: y = 91284

8 0
2 years ago
Read 2 more answers
(a) What are the possible values of l for n = 4? (Enter your answers as a comma-separated list.)
vodka [1.7K]

Answer:

(a) 0,1,2,3      (b) -3,-2,-1,0,1,2,3              (c) 6               (d) 5

Explanation:

(a) for the principal quantum number 'n', the possible values of I = 0 to n-1. Thus, if the principal quantum number 'n' =4, I = 0,1,2,3.

(b) for a given number of 'I', the possible values of ml = -I to +I. Therefore, if I =3, then ml = -3,-2,-1,0,1,2,3

(c) 'I' which is the orbital angular momentum quantum number usually has values from 0,1,2,⋯,n−1. Therefore, for n greater than or 6, t would be greater than or equal to 5. Thus, the smallest possible value of n for which I can be 6 is 6.

(d) In a 3-dimensional figure,  If the z-component of the orbital angular momentum Lz for which I=5 is measured, The possible outcomes will be:

mħ = -5ħ, -4ħ, -3ħ, -2ħ, -1ħ, 0, 1ħ, 2ħ, 3ħ, 4ħ, 5ħ.

Thus, the smallest possible l that can have a z component of 5ℏ is 5.

4 0
2 years ago
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