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3 years ago

How does insulin bind to cells and the mechanism involved in triggering the cells to take in glucose

Chemistry

1 answer:

Lera25 [3.4K]3 years ago

7 0

Explanation:

Once blood glucose levels increase, pancreatic insulin migrates into a fat cell via the blood stream. Insulin then binds in the plasma membrane of the cell to an Insulin Receptor (IR). Through autophosphorylation, phosphate groups are then added to the IR, causing GLUT4 molecules to come to the cell's surface.

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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

Suppose there are two known compounds containing the generic elements X and Y. You have a 1.00-g sample of each compound. One sa

Lapatulllka [165]

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason:

If you look at the combining masses of X and Y in each of the two compounds,

The first compound contains 0.25g of X combined with 0.75g of Y
so the ratio (by mass) of X to Y = 1 : 3

The second compound contains 0.33 g of X combined with 0.67 g of Y
so the ratio (by mass) of X to Y = 1 : 2

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!)

The first compound will then contain 4g of X and 12g of Y
The second compound will then contain 6g of X and 12g of Y

<span>The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6
<span>or 2 : 3 </span>

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 </span>

The two compounds given with the plausible formula must therefore contain the same ratio.

8 0

3 years ago

Cations are formed by _____. gaining protons losing protons gaining electrons losing electrons

Ede4ka [16]

Cations are formed by losing electrons in the valence shell of atoms. There are more protons than electron ions making the atom more positive.

7 0

3 years ago

Read 2 more answers

4. Fill in the blanks. Complete each sentence using a term from the word bank. (4)

skelet666 [1.2K]

Answer:

A.compound genuinely bozo L XL so

4 0

2 years ago

Calculate the ph of 0,24 m of kch3coo.?

Galina-37 [17]

Answer:

Correct option is A)

KaC

1.8×10

−6

=1.34×10

−3

pH=−log[H

]

=2.88

Explanation:

here is your answer if you like my answer please follow

6 0

3 years ago