in their mechanism of action, a difference between lipid-soluble and water-soluble hormones is that _____.

Answers

Answer 1

Answer: In their mechanism of action, a difference between lipid-soluble and water-soluble hormones is that lipid-soluble hormones diffuse into the cell's plasma membrane to access protein receptors, while water-soluble hormones bind to protein receptors on the plasma membrane without diffusion.

Examples of lipid-soluble hormones are steroids. Examples of water-soluble hormones are polypeptides.

Answer 2

Answer:

Final answer:

Lipid-soluble hormones pass through the cell membrane and directly interact with the cell's DNA, while water-soluble hormones bind to surface receptors, triggering intracellular signaling pathways.

Explanation:

In the field of Biology, the principal difference between lipid-soluble and water-soluble hormones lies in their mechanism of action. Lipid-soluble hormones, like steroid hormones, can easily pass through the cell membrane and interact directly with the cell's DNA to influence gene transcription. On the other hand, water-soluble hormones cannot cross the cell membrane due to their polarity. Instead, these hormones (such as peptides and amines) bind to receptors located on the cell's surface, triggering a series of intracellular events or signaling pathways that lead to a cellular response.In contrast, water-soluble hormones cannot passively cross the cell membrane due to their hydrophilic nature. They bind to specific receptors on the cell surface, triggering a cascade of intracellular signaling events through secondary messengers, such as cAMP or calcium ions. These signaling pathways ultimately result in the desired cellular response.

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Answers

Answer:

a) 13.2 moles $2H_(2)O$

b) 79.33 grams of $2H_(2)O$

Explanation:

First, we'll need to balance the equation

$H_(2(g)) + O_(2(g))$ → $H_(2)O_((g))$

There are 2 (O) on the left and only one on the right, so we'll add a 2 coefficient to the right.

$H_(2(g)) + O_(2(g))$ → $2H_(2)O_((g))$

Now there are 4 (H) on the right and only 2 on the left, so we'll add a 2 coefficient to the ( $H_(2)$ ) on the left.

$2H_(2(g)) + O_(2(g))$ → $2H_(2)O_((g))$

The equation is now balanced.

a) This can be solved with a simple mole ratio.

4.6 moles $O_(2)$ × $(2 moles H_(2)O)/(1 mole O_(2))$ = 13.2 moles $2H_(2)O$

b) This problem is solved the same way!

2.2 moles $H_(2)$ × $(2 moles H_(2)O)/(2 moles H_(2))$ = 2.2 moles $2H_(2)O$

However, this problem wants the mass of $2H_(2)O$ , not the moles.

The molecular weight of $2H_(2)O$ is the weight of 4 (H) molecules and 2 (O) molecules (found on the periodic table). So,

4(1.008) + 2(15.999) = 36.03 g/mol

2.2 moles $2H_(2)O$ × $(36.03 g)/(1 mol)$ = 79.33 grams of $2H_(2)O$

A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L flask and is found to contain 8.62 g of CO, 2.60 g of H2, 43.0 g of CH4, and 48.4 g of H2O.

Answers

The given question is incomplete. But the complete question is this:

A mixture of $CH_(4)$ and $H_2O$ is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L flask and is found to contain 8.62 g of CO, 2.60 g of $H_2$ , 43.0 g of $CH_(4)$ , and 48.4 g of $H_(2)O$ . Assuming that equilibrium has been reached, calculate $K_(p)$ for the reaction.