A) a disk formed of long trails of stars coiled into a spiral
<span>E = ħc / λ
ħ = plancks constant = 6.626x10^-34 Js
c = speed of light = 2.999x10^8 m/s
λ = wavelength of light = 670.8x10^-9 m
E = (6.626x10^-34 Js) x (2.999x10^8 m/s) x (1 / 670.8x10^-9 m)
E = 2.962x10^-19 J
</span><span>3x10^8 / (670.8 * 10^-9)
=4.47x10^14 Hz
4.47x10^14 Hz multiplied by plank's constant = 2.9634x10^-19
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The Lewis structure of P₄ is shown in 3-D form. The two bottom corner P atoms are facing right in front of us, one P atom behind the two, and one P above it. Each line represents 2 electrons. When you add the lone electrons, you get a total of 20 valence electrons.
Formal charge of each P: 5 - (2 +1/2*6) = 0
Answer:
CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Explanation:
To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:
pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2
Also, Kw / Kb = Ka
Thus:
pKa of CH3CH2NH3+/CH3CH2NH2 is:
Kw / kb = Ka = 1.79x10⁻¹¹
-log Ka = pKa
pKa = 10.75
pKa of C6H5NH3+/ C6H5NH2 is:
Kw / kb = Ka = 2.5x10⁻⁵
-log Ka = pKa
pKa = 4.6
That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
