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
The answer to your question is the first one!
Answer:
at first i would like to tell which class
Explanation:
125000 is the answwr
Answer:
The balanced chemical equation: NH₃ + 2 HF → NH₄⁺ + HF₂⁻
Explanation:
According to the Brønsted–Lowry acid–base theory, the acid- base reaction is a type of chemical reaction between the acid and base to give a conjugate acid and a conjugate base.
In this reaction, a Brønsted–Lowry acid loses a proton to form a conjugate base. Whereas, a Brønsted–Lowry base accepts a proton to form a conjugate acid.
Acid + Base ⇌ Conjugate Base + Conjugate Acid
The acid dissociation constant (Kₐ) <em>signifies the acidic strength of a chemical species.</em>
∵ pKₐ = - log Kₐ
Thus for a strong acid, Kₐ value is large and pKₐ value is small.
pKₐ (HF) = 3.2 → strong acid
pKₐ (NH₃) = 38 → weak acid
<u>The chemical reaction involved in the dissolution process:</u>
NH₃ + 2 HF → NH₄⁺ + HF₂⁻
In this acid-base reaction, the acid HF reacts with NH₃ base to give the conjugate base HF₂⁻ and conjugate acid NH₄⁺.
<u>HF (acid) donates a proton to form the conjugate base, HF₂⁻ ion. NH₃ (base) accepts a proton to form the conjugate acid. </u>

<u>Difference </u><u>between </u><u>Atomic </u><u>mass</u><u>, </u><u>relative </u><u>atomic </u><u>mass </u><u>and </u><u>average </u><u>atomic </u><u>mass</u><u> </u><u>:</u><u>-</u>
<h3><u>Atomic </u><u>Mass </u><u>:</u><u>-</u></h3>
- Atomic mass is the mass of neutrons and protons present in the nucleus of an atom .
- It is always calculated for a single element and having direct value
- For isotopes also, the atomic mass is calculated separately . Example :- <u>Carbon </u><u>1</u><u>2</u><u> </u><u>,</u><u> </u><u>carbon </u><u>1</u><u>3</u><u> </u><u>and </u><u>carbon </u><u>1</u><u>4</u><u> </u><u>have </u><u>different </u><u>atomic </u><u>mass</u><u>. </u>
- The SI unit of Atomic mass is " u" and "amu"
<h3>
<u>Relative </u><u>Atomic </u><u>mass </u><u>:</u><u>-</u></h3>
- Relative atomic mass is mean mass of the atoms of an element which is compared to the 1/12th mass of carbon - 12 .
- Carbon - 12 is taken as a relative when we calculate the relative atomic mass of any element
- For calculating relative atomic mass, we need to know the masses, percentage and abundance of all types of elements
- Relative atomic mass is a dimension less quantity
<h3><u>Average </u><u>Atomic </u><u>Mass </u><u>:</u><u>-</u></h3>
- Average atomic mass is the average mass of an atoms of a particular element by considering it's isotopes
- While we calculate average atomic mass is a standardized number. Whereas, Average atomic mass sometimes varies geologically .
- It also includes percentage, abundance and masses of given element .
- In average atomic mass, We do not compare mean value with the 1/12 mass of carbon - 12
- The unit of Average atomic mass is "Amu" or " u " .