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.
Answer:
As a substance changes from a solid to a liquid to a gas, its molecules first the molecules are moving fast enough, they are able to "escape." They leave the surface of the liquid as gas molecules. Evaporation is not the only process that can change a substance from a liquid to a gas. The same change can occur through boiling.
Explanation:
hopfully this helps!
Answer:
The change in internal energy is - 1.19 kJ
Explanation:
<u>Step 1:</u> Data given
Heat released = 3.5 kJ
Volume calorimeter = 0.200 L
Heat release results in a 7.32 °C
Temperature rise for the next experiment = 2.49 °C
<u>Step 2:</u> Calculate Ccalorimeter
Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C
Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C
<u>Step 3:</u> Calculate energy released
Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ
<u>Step 4:</u> Calculate change in internal energy
ΔU = Q + W W = 0 (no expansion)
Qreac = -Qcal = - 1.19 kJ
ΔU = - 1.19 kJ
The change in internal energy is - 1.19 kJ
Answer:
2Mg (s) + 
(g) ---> 2MgO (s)
Explanation:
Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
