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bezimeni [28]
2 years ago
5

In _____ , materials move from an area of higher concentration to an area of lower concentration through a cell membrane.

Chemistry
1 answer:
Elenna [48]2 years ago
4 0
Answer = 1.selective permeability
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Round to 4 significant figures.<br> 35.5450
Lelu [443]
Answer:
35.5450 will be rounded to 35.55
Explanation:
=35.5450
if the last digit is less than 5 then it will be ignored
=35.545
when the dropping digit is 5 then the retaining digit will increse by a factor of 1
=35.55
i hope this will help you
3 0
3 years ago
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ANSWER FAST PLEASE!<br> Balance the following equation:<br> __Hg + __O₂ --&gt; __HgO
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Answer:

4Hg+2O2=4HgO

four Mercury + four oxygen

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For every 1.0 mole of glycine in the sample, how many molecules of methionine are present? (for help performing calculations wit
krok68 [10]

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3 years ago
I need help ASAP!!!!! What happens to water when it changes to ice?
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6 0
3 years ago
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Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
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