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Mekhanik [1.2K]
3 years ago
15

4

Chemistry
2 answers:
saveliy_v [14]3 years ago
7 0

Answer:

Is A- Carbon Dioxide

Explanation:

other air pollutants collect in the atmosphere and absorb sunlight and solar radiation that have bounced off the earth's surface.

Llana [10]3 years ago
3 0

Answer:Carbon dioxide would be your best choice.

Explanation:

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If the Kelvin temperature of a gas is doubled, the volume of the gas will increase by ____
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If the Kelvin temperature of a gas is doubled, the volume of the gas will increase by two. It follows Charles law where in for a mixed gas of mass, the volume is directly proportional to the temperature at constant pressure.

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1. At 7 a.m., the science classroom had a temperature of 68°F. By 3 p.m., the temperature had
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What is it like on Mars? Please type 6 good answers.
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3 years ago
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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
kumpel [21]

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

P_{50} is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

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

P_{50} = 22 mm Hg

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

Hence, the fractional saturation for hemoglobin is 0.86.

4 0
3 years ago
This balanced chemical equation represents a chemical reaction: 6no + 4nh3 → 5n2 + 6h2o what volume of nh3 gas, at standard temp
Liula [17]

The answer is: volume of ammonia gas is 7.4 L.

Chemical reaction: 6NO + 4NH₃ → 5N₂ + 6H₂O.

m(NO) = 15 g; mass of nitrogen(II) oxide.

M(NO) = 30 g/mol; molar mass of nitrogen(II) oxide.

V(NH₃) = ?

n(NO) = 15 g ÷ 30 g/mol.

n(NO) = 0.5 mol; amount of nitrogen(II) oxide.

From chemical reaction: n(NO) : n(NH₃) = 6 : 4.

0.5 mol : n(NH₃) = 6 : 4.

n(NH₃) = 0.33 mol; amount of ammonia.

Vm = 22.4 L/mol; molar volume at STP.

V(NH₃) = 0.33 mol · 22.4 L/mol..

V(NH₃) = 7.4 L.

3 0
4 years ago
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