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Neporo4naja [7]
3 years ago
13

4. How many moles of KBr are found in 3 Liters of 0.4 M solution?

Chemistry
1 answer:
Allisa [31]3 years ago
4 0

There are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.

<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>

The number of moles of a substance can be calculated by multiplying the molarity by the volume.

No. of moles = Molarity × volume

According to this question, 3L of a KBr solution are contained in a 0.4M.

no. of moles = 3L × 0.4M = 1.2moles

Therefore, there are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.

Learn more about no. of moles at: brainly.com/question/14919968

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Neporo4naja [7]
If its a unicellular orgnism, it has only one cell . no?
so i guess the answer has to be 1

3 0
3 years ago
Hi! I really really need help roght now please help me and if u put whatever in the chas as in just to get points i will ban you
dolphi86 [110]

The correct order given below shows the changes that occurs in a mice population in response to changes in their environment:

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<h3>What is the correct order for natural selection in the desert environment given?</h3>

Based on the process of natural selection due to envrionmental pressures, the population of the mice in the desert changes as follows before and after the environmental change:

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Therefore, the correct order shows the changes that occurs in a mice population in response to changes in their environment.

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8 0
3 years ago
The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}&#10;

Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s&#10;}{K_{M} ln (\frac{45 mmol/L&#10;}{9 mmol/L&#10;}) + (45 mmol/L - 9 mmol/L&#10;)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

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