Ask question

Ask question

All categories

3 years ago

12

Concentrated hydrochloric acid, HCl, comes with an approximate molar concentration of 12.1 M. If you are instructed to prepare 3

50.0 mL of a 0.975 M HCl solution, how many milliliters of the stock (concentrated) HCl solution will you use

1 answer:

Semmy [17]3 years ago

6 0

Answer:

28.20 mL of the stock solution.

Explanation:

Data obtained from the question include the following:

Molarity of stock solution (M1) = 12.1 M

Volume of diluted solution (V2) = 350.0 mL

Molarity of diluted solution (M2) = 0.975 M

Volume of stock solution needed (V1) =..?

The volume of stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12.1 x V1 = 0.975 x 350

Divide both side by 12.1

V1 = (0.975 x 350)/12.1

V1 = 28.20 mL.

Therefore, 28.20 mL of the stock solution will be needed to prepare 350.0 mL of 0.975 M HCl solution.

You might be interested in

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

7 0

3 years ago

What did you include in your response? Check all that apply.

Aliun [14]

Answer:

Without corals, the algae are not protected and cannot perform photosynthesis.

5 0

3 years ago

Read 2 more answers

What's the Balance equation for:<br>_C4H10O +_ O2--->_CO2+_H20

steposvetlana [31]

Answer:

C₄H₁₀O + 6O₂ ⇒ 4CO₂ + 5H₂O

Explanation:

Match the amount of reactants and products on both sides of the equation.

3 0

3 years ago

23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp

stepan [7]

Answer:- 13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

$P_1V_1=P_2V_2$

From given information:-

$P_1$ = 58.7 Kpa

$V_1$ = 23.5 L

$P_2$ = 101.325 Kpa

$V_2$ = ?

Let's plug in the values and solve it for final volume.

$58.7Kpa*23.5L=101.325Kpa*V_2$

On rearranging the equation for $V_2$

$V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}$

$V_2$ = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

3 0

3 years ago

Which option correctly describes the enthalpy of formation?

leonid [27]

Answer: Endothermic reaction.

Explanation: A chemical reaction in which energy is absorbed from the surrounding environment.

4 0

3 years ago