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Sagot :
Answer:
1.497 mol Ir
General Formulas and Concepts:
Math
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Chemistry
Atomic Structure
- Reading a Periodic Table
Stoichiometry
- Using Dimensional Analysis
Explanation:
Step 1: Define
287.7 g Ir
Step 2: Identify Conversions
[PT] Molar Mass of Ir - 192.22 g/mol
Step 3: Convert
- Set up: [tex]\displaystyle 287.7 \ g \ Ir(\frac{1 \ mol \ Ir}{192.22 \ g \ Ir})[/tex]
- Multiply/Divide: [tex]\displaystyle 1.49672 \ mol \ Ir[/tex]
Step 4: Check
Follow sig fig rules and round. We are given 4 sig figs.
1.49672 mol Ir ≈ 1.497 mol Ir
Answer:
[tex]\boxed {\boxed {\sf About \ 1.497 \ mol \ Ir }}[/tex]
Explanation:
To convert from grams to moles, we must the molar mass of iridium, which can be found on the Periodic Table.
- Ir: 192.22 g/mol
Use the molar mass as a ratio.
[tex]\frac{ 192.22 \ g \ Ir}{ 1 \ mol \ Ir}[/tex]
Multiply by the given number of grams: 287.7
[tex]287.7 \ g Ir *\frac{ 192.22 \ g \ Ir}{ 1 \ mol \ Ir}[/tex]
Flip the fraction so the grams of Iridium will cancel.
[tex]287.7 \ g Ir *\frac{1 \ mol \ Ir }{ 192.22 \ g \ Ir}[/tex]
[tex]287.7 *\frac{1 \ mol \ Ir }{ 192.22 }[/tex]
[tex]\frac {287.7 \ mol \ Ir }{192.22} \\[/tex]
Divide.
[tex]1.49672251 \ mol \ Ir[/tex]
The original measurement has 4 significant figures, so we must round our answer to 4 sig figs as well. For the answer we calculated, that is the thousandth place.
The 7 in the ten thousandth place tells us to round the 6 up to a 7.
[tex]1.497 \ mol \ Ir[/tex]
There are about 1.497 moles of iridium in 287.7 grams.
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