Answer
2.7956 * 10^19 photons
Givens
- Wavelength = λ = 525 * 10^-9 meters [1 nmeter = 1*10^-9 meters]
- c = 3 * 10^8 meters
- E = ???
- W = 100 watts
- t = 1 second
- h= plank's Constant = 6.26 * 10^-34 J*s
Formula
E = h * c / λ
W = E / t
Solution
E = 6.26 * 10^-34 j*s * 3 * 10^8 m/s /525 * 10^-9 (m)
The meters cancel out. So do the seconds. You are left with Joules as you should be.
E = 3.577 * 10^-18 Joules
What you have found is the energy of 1 photon.
Now you have to find the Joules from the watts.
W = E/t
100 * 1 second = 100 joules
1 photon contains 3.577 * 10 ^ - 18 Joules
x photon = 100 joules
1/x = 3.577 * 10^-18 / 100 Cross multiply
100 = 3.577 * 10 ^ - 18 * x Divide both sides by 3.577 * 10 ^ - 18
100/3.577 * 10 ^ - 18 = 3.577 * 10 ^ - 18x / 3.577 * 10 ^ - 18
2.7956 * 10^19 photons = x
Answer:
In the center is the eye, with nearly clear skies, surrounded by the violent eyewall, with the strongest winds and very heavy rain.
Explanation:
<span> A homogeneous mixture is composed of a single visible phase while a heterogeneous mixture has two or more</span>
Answer:
Explanation:
Hola,
En este caso, podemos usar la ley de Boyle, la cual nos permite analizar el comportamiento volumen-presión en un gas ideal de manera inversamente propocional:
Así, dado el volumen y la presión inicial, la cual se convierte a atmósferas (760 torr = 1atm), calculamos el volumen final a 1 atm como se muestra a continuación:
Saludos!
One of the best buffer choice for pH = 8.0 is Tris with Ka value of 6.3 x 10^-9.
To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value:
pKa = -log Ka
For Tris, which is an abbreviation for 2-Amino-2-hydroxymethyl-propane-1,3 -diol and has a Ka value of 6.3 x 10^-9, the pKa is
pKa = -log Ka
= -log (6.3x10^-9)
= 8.2
We know that buffers work best when pH is equal to pKa:
pKa = 8.2 = pH
Therefore Tris would be a best buffer at pH = 8.0.
