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Whitepunk [10]
3 years ago
9

Identify the functional group(s) that appear in betaxolol. This compound is in a class of drugs called beta-blockers, which are

used to lower blood pressure, lower heart rate, reduce angina (chest pain), and reduce the risk of recurrent heart attacks Alcohol Ether Arene Carboxylic Acid Aldehyde Ester Amine Alkene KetoneFigure:contains some chemical structures

Chemistry
1 answer:
Lady_Fox [76]3 years ago
5 0

Answer:

  1. Alcohol
  2. Ether
  3. Arene
  4. Amine

Explanation:

In the attached picture you may find the structure of betaxolol.

You can see the alcohol group C-O-H as well as the ether group C-O-C.

The arene -or aromatic ring- can also be seen.

There's also a secondary amine group, C-NH-C.

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Consider the reaction: CO2(g) + CCl4(g) ⇌ 2 COCl2(g) ΔG° = 46.9 kJ Under the following conditions at 25 oC: LaTeX: P_{CO_2}P C O
Mamont248 [21]

Answer : The value of \Delta G_{rxn} is, -47.0kJ/mole

Explanation :

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln K_p   ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction

\Delta G_^o =  standard Gibbs free energy  = 46.9 kJ

R = gas constant = 8.314 J/mole.K

T = temperature = 25^oC=273+25=298K

K_p = equilibrium constant

First we have to calculate the value of K_p.

The given balanced chemical reaction is,

CO_2(g)+CCl_4(g)\rightarrow 2COCl_2(g)

The expression for reaction quotient will be :

K_p=\frac{(p_{COCl_2})^2}{(p_{CO_2})\times (p_{CCl_4})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

K_p=\frac{(0.653)^2}{(0.459)\times (0.984)}

K_p=0.944

Now we have to calculate the value of \Delta G_{rxn} by using relation (1).

\Delta G_{rxn}=\Delta G^o+RT\ln K_p

Now put all the given values in this formula, we get:

\Delta G_{rxn}=-46.9kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (0.944)

\Delta G_{rxn}=-47.0kJ/mol

Therefore, the value of \Delta G_{rxn} is, -47.0kJ/mole

3 0
3 years ago
Show a numerical setup for calculating the number of moles in the sample of CaCO3
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The numerical setup is

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gCO2 cancels out so it is

11 x 1 mol CO2/44.1 CO2= 0.25 mol CO2

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see the attached file!!!

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