<h3>
Answer:</h3>
91.2 g Mn
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 1.00 × 10²⁴ atoms Mn
<u>Step 2: Identify Conversions</u>
Avogadro's Numer
[PT] Molar Mass of Mn - 54.94 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
91.2321 g Mn ≈ 91.2 g Mn
These problems are a bit interesting. :)
First let's write the molecular formula for ammonium carbonate.
NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)
17.6 gNH4CO3
Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.
17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)
Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)
NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol
17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4
Now just take the molar mass we found to convert that amount into moles!
4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4
E=hc/l
E=
<span><span>E=<span>(6.626 x 10-34 J s)(3.0 x 108m/s )</span><span>=2.88 x 10-19J</span></span><span>6.90 x 10-7m</span></span>
A. Gamma radiation because it reaches further than beta and alpha rays
