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Sagot :
Answer:
0.127 mol Au
General Formulas and Concepts:
Math
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Chemistry
Atomic Structure
- Reading a Periodic Table
- Moles
Stoichiometry
- Using Dimensional Analysis
Explanation:
Step 1: Define
[Given] 25.0 g Au
[Solve] moles Au
Step 2: Identify Conversions
[PT] Molar Mass of Au - 196.97 g/mol
Step 3: Convert
- [DA] Set up: [tex]\displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})[/tex]
- [DA] Multiply/Divide [Cancel out units: [tex]\displaystyle 0.126923 \ mol \ Au[/tex]
Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
0.126923 mol Au ≈ 0.127 mol Au
Answer:
[tex]\boxed {\boxed {\sf 0.127 \ mol \ Au}}[/tex]
Explanation:
To convert from mass to moles, the molar mass must be used. This can be found on the Periodic Table.
- Gold (Au): 196.96657 g/mol
Use this number as a ratio.
[tex]\frac {196.96657 \ g\ Au }{1 \ mol \ Au}[/tex]
Multiply by the given number of grams.
[tex]25.0 \ g \ Au*\frac {196.96657 \ g\ Au }{1 \ mol \ Au}[/tex]
Flip the fraction so the grams of gold will cancel.
[tex]25.0 \ g \ Au*\frac {1 \ mol \ Au}{196.96657 \ g\ Au }[/tex]
[tex]25.0*\frac {1 \ mol \ Au}{196.96657 }[/tex]
[tex]\frac {25.0 \ mol \ Au}{196.96657 }[/tex]
[tex]0.1269250919 \ mol \ Au[/tex]
The original measurement has 3 significant figures, so the answer must have the same. For the number we calculated that is the thousandth place. The 9 in the ten thousandth place tells us to round the 6 to a 7.
[tex]0.127 \ mol \ Au[/tex]
25.0 grams of gold is about 0.127 moles of gold.
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