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Nesterboy [21]
3 years ago
8

(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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2 years ago
Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is a
musickatia [10]

Answer:

Approximately 1.876 \times 10^{-3}\; \rm mol.

Explanation:

Convert both volumes to standard units (that is: liters.)

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Number of moles of \rm HCl initially present (in the 10.00\; \rm mL solution at 1.00\; \rm M.)

n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol.

Number of moles of \rm NaOH from the titration:

n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol.

\rm NaOH neutralizes \rm HCl at a 1:1 ratio:

\rm HCl + NaOH \to NaCl + H_2O.

Hence, n(\mathrm{HCl},\, \text{leftover}) = n(\mathrm{NaOH}) = 8.124 \times 10^{-3}\; \rm mol.

\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}.

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