Express each quantity in regular notation along with its appropriate unit. a. 3.60 × 1 0 5 s 3.60×10 5 s b. 5.4 × 1 0 − 5 g / c
1 answer:
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Answer:
9.430 * 10¹⁷ protons per second whill shine on the book from a 62 W bulb
Explanation:
To answer this question, first let's calculate the energy of a single photon with a wavelength (λ) of 504 nm:
E = hc/λ
Where h is Planck's constant (6.626*10⁻³⁴ J·s) and c is the speed of light (3*10⁸ m/s).
E = 6.626*10⁻³⁴ J·s * 3*10⁸ m/s ÷ (504*10⁻⁹m) = 3.944 * 10⁻¹⁹ J.
So now we can make the equivalency for this problem, that
<u>1 proton = 3.944 * 10⁻¹⁹ J</u>
Now we convert watts from J/s to proton/s:
1 = 1 W
Solving the problem, a 62 W bulb converts 5% of its output into light, so:
- 62 * 5/100 = 3.1 W
3.1 watts are equal to [ 2.535*10¹⁸ proton/s * 3.1 ] = 7.858 * 10¹⁸ proton/s
Of those protons per second, 12% will shine on the chemistry textbook, thus:
7.858 * 10¹⁸ proton/s * 12/100 = 9.430 * 10¹⁷ protons/s
<h3>Answer:</h3>
7.4797 g AlF₃
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
<u>Step 1: Define</u>
[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃
[Given] 15.524 g KF
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol AlF₃ = 6 mol KF
Molar Mass of Al - 26.98 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of K - 39.10 g/mol
Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol
Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
7.47966 g AlF₃ ≈ 7.4797 g AlF₃