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3 years ago

(20 Pt.s) Guys! You have to explain:

Chemistry

1 answer:

Nadya [2.5K]3 years ago

3 0

Weakly basic drugs behaves different from acidic drugs which is discussed below.

<h3><u>Explanation</u>:</h3>

Weakly basic drugs are those drugs which have an amine group associated with them. They are able to gain a proton to be come positively charged.

So drugs like quinine, ephedrine and aminopyrine which are basic got completely ionised in stomach.

The stomach can absorb those compounds which are lipid soluble. The acidic drugs like alcohols, Salicylic acid, aspirin, thiopental, secobarbital and antipyrine etc which are acidic gets absorbed by means of <em>diffusion</em> through the membrane.

But the basic drugs have charges on them which makes them lipophobic. So they cannot get absorbed through stomach. However weakly basic drugs sometimes get absorbed depending on their ionisation extent.

The rest goes to small intestine which has basic environment and there they gets absorbed via diffusion or facilitated diffusion.

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A 44% wt/wt solution of H2SO4 has a density of 1.343g/ml. What mass of H2SO4 is 60ml of this solution?

horsena [70]

<u>Answer:</u> The mass of sulfuric acid present in 60 mL of solution is 34.1 grams

<u>Explanation:</u>

We are given:

44 % (m/m) solution of sulfuric acid. This means that 44 grams of sulfuric acid is present in 100 grams of solution.

To calculate volume of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.343 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

$1.343g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.343g/mL}=77.46mL$

To calculate the mass of sulfuric acid present in 60 mL of solution, we use unitary method:

In 77.46 mL of solution, mass of sulfuric acid present is 44 g

So, in 60 mL of solution, mass of sulfuric acid present will be = $\frac{44g}{77.46mL}\times 60mL=34.1g$

Hence, the mass of sulfuric acid present in 60 mL of solution is 34.1 grams

3 0

3 years ago

For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t

Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) $2.136197(10^{25})$ Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = $6.02.(10^{23})$ elemental units

For example :

1 mol of oxygen atoms = $6.02.(10^{23})$ oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

$molarmass_{(Al)}=26.982\frac{g}{mol}$

$molarmass_{(S)}=32.065 \frac{g}{mol}$

$molarmass_{(O)}=15.999\frac{g}{mol}$

If we sum all the molar masses = $(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}$

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

$2.663 kg=2663g$

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

$x=\frac{2663}{75.046}mol=35.485 mol$

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = $6.02(10^{23})$

35.485 moles of Al have $(35.485)(6.02)(10^{23})=2.136197(10^{25})$ Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

$x=\frac{(2663).(15.999)}{75.046} g$

x = 567.723 g of O

6 0

3 years ago

Someone help me with this ASAP

ZanzabumX [31]

Answer:

First bone

Explanation:

6 0

4 years ago

Read 2 more answers

Synthetic, or man-made, elements typically exist for only a short time

o-na [289]

This is true. Elements past lead are radioactive, because the repulsive force of the protons cannot be overpowered by the “gluing” ability of neutrons (remember, likes repel). As more and more protons are added, generally, the elements become more unstable; for example, Bismuth, right next to lead on the Periodic Table, is radioactive, but the half life of this element is about a billion times longer than the current age of the universe, but Oganesson, element number 118, has a half life of fractions of a second.

8 0

3 years ago

What product(s) forms at the cathode in the electrolysis of an aqueous solution of k2so4?

fenix001 [56]

Catod(-)
K⁺
2H20 +2e⁻ ---> H2 +2OH⁻

We can also say, that
K⁺ +OH⁻ +H2 = KOH +H2
At the cathode KOH and H2 are formed.

8 0

3 years ago