A student uses visible spectrophotometry to determine the concentration of CoCl2(aq) in a sample solution. First the student prepares a set of CoCl2(aq) solutions of known concentration. Then the student uses a spectrophotometer to determine the absorbance of each of the standard solutions at a wavelength of 510nm and constructs a standard curve. Finally, the student determines the absorbance of the sample of unknown concentration. A wavelength of 510nm corresponds to an approximate frequency of 6×1014s−1 . What is the approximate energy of one photon of this light? 9×1047J 9×1047J A 3×1017J 3×1017J B 5×10−7J 5×10−7J C 4×10−19J 4×10−19J D Submit

3.97 x 10^-19 J

Explanation:

The energy E of a photon is given by:

       E = hf where h = Planck's constant = 4.14 × 10−15 eV · s and f = frequency

In this case, f = 6×10^14s−1, hence, the energy E of one photon of the light would be;

E = 4.14 × 10−15 eV · s x 6×10^14s−1

               =  2.484 eV

Note that:

1 eV = 1.60 × 10−19 J

Hence,

2.484 eV = 2.484 x 1.60 × 10−19 J

              = 3.97 x 10^-19 J

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j