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svp [43]
3 years ago
5

How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work

used to solve this problem.
Chemistry
1 answer:
strojnjashka [21]3 years ago
8 0
2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!
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An inhibitor is added to an enzyme-catalyzed reaction at a concentration of 26.7 μM. The Vmax remains constant at 50.0 μM/s, but
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Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

<h3>What is the Ki for the inhibitor?</h3>

The Ki of an inhibitor is known as the inhibition constant.

The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.

Using the Michaelis-Menten equation for inhibition:

  • Kma = Km/(1 + [I]/Ki)

Making Ki subject of the formula:

  • Ki = [I]/{(Kma/Km) - 1}

where:

  • Kma is the apparent Km due to inhibitor
  • Km is the Km of the enzyme-catalyzed reaction
  • [I] is the concentration of the inhibitor

Solving for Ki:

where

[I] = 26.7 μM

Km = 1.0

Kma = (150% × 1 ) + 1 = 2.5

Ki = 26.7 μM/{(2.5/1) - 1)

Ki = 53.4 μM

Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

Learn more about enzyme inhibition at: brainly.com/question/13618533

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The standard cell potential of the following galvanic cell is 1.562 V at 298 K. Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s) What is the
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Answer:

E = 1.602v

Explanation:

Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …

             Zn⁰(s) => Zn⁺²(aq) + 2 eˉ

2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)          

_____________________________

Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)

Given E⁰ = 1.562v

Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044

E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v

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