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3 years ago

Which electrolyte is used in an alkali fuel cell?

Chemistry

1 answer:

lyudmila [28]3 years ago

8 0

Answer: an aqueous potassium hydroxide solution

An alkaline fuel cell is a zero-emission device whose one major component is the electrolyte. An electrolyte, on the other hand, is a solution that is able to conduct electricity. In an alkaline fuel cell. The electrolyte is an alkaline liquid, and potassium hydroxide also known as KOH is the only alkaline liquid among the choices.

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How many atoms are in an oxygen molecule (O 2 ) ?

saw5 [17]

Answer:

Explanation:

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3 years ago

Which isotope of nirtogen is more abundent?

Jobisdone [24]

<span>Nitrogen-14 should be the correct response.</span>

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4 years ago

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.

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3 years ago

What else should you listen for as a speaker gives a speech?

kenny6666 [7]

Answer:

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Explanation:

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2 years ago

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You’d need to mix over have the mass of the 450 gram solution

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