Stoichiometry is not really hard if you are good with your numbers. All you have to do is pay close attention and follow you teacher because if you don't you will end up like me not knowing what you are doing. Listen and pay real close attention and I can guarantee you it will be just fine!
In 0.190 mole of C6H14O, there is 0.190*6 (number of C in one molecule) = 1.140 mole of C atoms. The total number of C atoms = 1.14 * 6*
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(atoms of C in one mole) = 6.84*
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atoms.
Answer:
A is your answer..... !!!!!!!!!!!!!
<h3>
Answer:</h3>
0.819 mol Ag
<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>
- 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>
4.93 × 10²³ atoms Ag
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
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- Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.818665 mol Ag ≈ 0.819 mol Ag
Answer:
3
Explanation:
3 is the answer is did this a little while ago