To calculate how many photons are in a certain amount of energy (joules) we need to know how much energy is in one photon.
Start by using two equations:
Energy of a photon = Frequency * Planck's constant (6.626 * 10^(-34) J-s)
Speed of light (constant 3 * 10^8 m/s) = Frequency * Wavelength
Which means:
frequency = Speed of Light / Wavelength
So energy of a photon = (Speed of light * Planck's constant)/(Wavelength)
You may have seen this equation as E = hc/<span>λ</span>
We have a wavelength of 691 nm or 691 * 10^-9 meters
So we can plug in all of our knowns:
E = (6.626 * 10^(-34) J-s) * (3.00 * 10^8 m/s) / (691 * 10^-9 m) =
2.88 * 10^(-19) joules per photon
Now we have joules per photon, and the total number of joules (0.862 joules)
,so divide joules by joules per photon, and we have the number of photons:
0.862 J/ (2.88 * 10^(-19) J/photon) = 3.00 * 10^18 photons.
KI-starch paper allows the detection of strong oxidizers such as nitrite. It is used here to control diazotization of 4-nitroaniline. Nitrite oxidizes potassium iodide in order to form elemental iodine which reacts with starch to a blue-violet complex. With KI-starch paper, enough sodium nitrite is added to produce nitrous acid, which <span>then will react with 4-nitroaniline to form a diazonium salt.</span>
Answer:

Explanation:
Given


Required
Determine the percentage error
First, we need to determine the difference in the measurement




The percentage error is calculated as thus:




<em>approximated</em>
The total volume of water that would be removed will be 75 mL
<h3>Dilution equation</h3>
Using the dilution equation:
M1V1 = M2V2
In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M
Substitute:
V2 = 500 x 10.20/12
= 425 mL
The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:
500 - 425 = 75 mL
More on dilution can be found here: brainly.com/question/7208939
I'm not sure how many sign fig's you are required to have.
However I think the final answer would be 0.05 Moles, because of the .5g, that is considered 1 sign fig.