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Serjik [45]
3 years ago
5

Which is an example of a condition in which homeostasis is maintained using a negative feedback loop?

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
7 0
A. low blood glucose <span>is an example of a condition in which homeostasis is maintained using a negative feedback loop</span>
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At 700 K, the reaction 2SO2(g) + O2(g) &lt;====&gt; 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f
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Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

  • If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
  • If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
  • If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }

Q=\frac{10^{2} }{0.10^{2} *0.10}

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

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3 years ago
32P can be used to make any nucleotide (A, C, G, or T) radioactive. Which of the following explains why this is true?
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B. The answer is: All nucleotides have a phosphorus atom that can be replaced with 32P.

Nucleotides contain a nitrogenous base, a five-carbon sugar, and, at least, one phosphate group. Exactly that phosphate group in the nucleotide has the phosphorus atom. Therefore, the phosphorus atom in the nucleotide can be replaced with radioactive phosphorus-32 (32P). 
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3 years ago
While doing a lab a student found the density of a piece of pure aluminum to be 2.85 g/cm3 the accepted value for the density of
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Answer:

THE PERCENT ERROR IS 5.55 %

Explanation:

To calculate the percent error, we use the formula:

Percent error = Found value - accepted value / accepted value * 100

Found value = 2.85 g/cm3

Accepted value = 2.70 g/cm3

Solving for the percent error, we have:

Percent error = 2.85 g/cm3 - 2.70 g/cm3 / 2.70 g/cm3 * 100

Percent error = 0.15 / 2.70 * 100

Percent error = 0.05555 * 100

Percent error = 5.55 %

In conclusion, the percent error is 5.55 %

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Explanation:

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