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ch4aika [34]
3 years ago
10

Identifi the atomHe ONeNa​

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
7 0

Answer:i wanna say he

Explanation:

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LUCKY_DIMON [66]

Here's one! those spots are actually storms:)


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3 years ago
The diagram below represents a portion of a cell membrane. The arrow indicates that the cell membrane is carrying out the proces
Llana [10]

Answer:

I'm feeling nice today so heres the answer

Explanation:

In the portion of the cell membrane shown in the diagram, the arrow indicates the process of active transport.

Explanation:

Active transport is one of the mechanisms of transmembrane transport, which involves the use of energy. The diagram (see image) shows the hydrogen (H⁺) output from the cytoplasm to the extracellular space, through an H⁺ pump —consuming ATP— which represents an active transport process.

The hydrophobic nature of the cell membrane prevents the free passage of hydrosoluble elements or ions, as H⁺, so they require the use of active transport to pass through it.

The other options presented are not correct, because

Respiration is a process that occurs in the mitochondria.

Diffusion is a passive transport process that does not require energy.

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8 0
3 years ago
Plzzzzzzzzzzz help what is a font!
spayn [35]

Answer:

1. a receptacle in a church for the water used in baptism, typically a freestanding stone structure.

2. A type of writing or text style

Explanation:

There are mutiple definitions of font

6 0
3 years ago
electric current is passed through water, it decomposes inti hydrogen and oxygen. write full balanced chemical equation for the
Tems11 [23]

Answer:

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6 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}


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
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]

              = 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.

6 0
3 years ago
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