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Answer: 78.69 g of Fe
Explanation: Iron(III) oxide is Fe2O3. It has a molar mass of 159.7 g/mole. That consists of:
2 x Fe at 55.85 = 11.69 or 69.94%
3 x O at 16.0 = 48.0 or 30.06%
Therefore 112.5 grams will contain (112.5 g)*(69.94%) = 78.69 g of Fe
Answer:
Label A shows cold area on the mug, while label B shows a warm area.
Explanation:
Have a good day
Answer:
27.03 %
Explanation:
The formula for the calculation of moles is shown below:
Given: For calcium
Given mass = 10.0 g
Molar mass of calcium = 40.078 g/mol
<u>Moles of calcium = 10.0 g / 40.078 g/mol = 0.2495 moles</u>
According to the given reaction:
1 mole of calcium on reaction forms 1 mole of calcium hydroxide
Thus,
0.2495 moles of calcium on reaction forms 0.2495 mole of calcium hydroxide
<u>Moles of calcium hydroxide = 0.2495 moles</u>
Molar mass of calcium hydroxide = 74.093 g/mol
Thus, Mass = Moles * Molar mass = 0.2495 moles * 74.093 g/mol = 18.5 g
<u>Theoretical yield = 18.5 g</u>
<u>
Given experimental yield = 5.00 g
</u>
<u>% yield = (Experimental yield / Theoretical yield) × 100 = (5.00/18.5) × 100 = 27.03 %
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