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Reil [10]
3 years ago
10

"determine the mass of oxygen" in a 7.9 g sample of al2(so4)3.

Chemistry
1 answer:
jeyben [28]3 years ago
7 0

Answer:

              4.43 g of Oxygen

Explanation:

As shown in Chemical Formula, one mole of Aluminium Sulfate [Al₂(SO₄)₃] contains;

                          2 Moles of Aluminium

                          3 Moles of Sulfur

                          12 Moles of Oxygen

Also, the Molar Mass of Aluminium Sulfate is 342.15 g/mol. It means,

          342.15 g ( 1 mole) of Al₂(SO₄)₃ contains  =  192 g (12 mole) of O

So,

                         7.9 g of Al₂(SO₄)₃ will contain  =  X g of O

Solving for X,

                       X  =  (7.9 g × 192 g) ÷ 342.15 g

                      X =  4.43 g of Oxygen

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Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa
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Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Where Y_{O_{2} } \\ is the fractional oxygen saturation

{P_{O_{2} } is the partial pressure of oxygen

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and h is the Hill coefficient

From the question,

{P_{O_{2} } = 40 mm Hg

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h = 3

Putting these values into the equation, we get

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3}  }

Y_{O_{2} } = \frac{64000 }{10648 + 64000  }

Y_{O_{2} } = \frac{64000 }{74648 }

Y_{O_{2} } = 0.86

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