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lianna [129]
3 years ago
12

Is Acetaminophen ionic, molecular or organic compound

Chemistry
2 answers:
Gre4nikov [31]3 years ago
8 0

Answer: A molecular compound.

Acetaminophen is an interesting compound in that has at least 3 specific types of functional groups.

LOOK UP THE STRUCTURE FOR Acetaminophen. It will make it clearer.

Or,

Draw a Benzene ring. This is the archetypal “aromatic ring”. 6 carbons in a ring with 3 alternating unsaturations or “double bonds”. It is planar, and follows the 2n + 2 rule for aromaticity. (Where n = 2, so (2 x 2) + 2 = 6. (The three unsaturations; you may not have seen all this yet, but you will, so bear with me).

The 2nd functional group, a -OH or a Hydroxyl group, behaves VERY differently on an aromatic ring than on an alkane. The resonance stabilization (of the aromatic ring), allows the Hydrogen to disassociate from the -OH to yield Aromatic ring—O - + H+ when it is in a polar solution, like water. Phenol is the simplest example.

3rd functional group is a methyl Amide. This amide is —NH—COCH3 (trans). Amides are a N bonded to a H and a carbon bearing a double bonded Oxygen, and the same carbon, in this example, a methyl group. There is a resonance between the N and O, but the structure you see is the lowest energy and most stable. Place the —OH Exactly opposite on the ring to the methyl amide, in the 1,4, or para position.

This may be overdoing it, but it never hurts to learn functional groups as soon as possible.

sergejj [24]3 years ago
3 0

Answer:

Organic compound

Explanation:

Acetominophen's molecular formula is C8H9NO2. Right from the formula you can also see that the compound has C (carbon) in it, therefore making it an organic compound (in which the definition is that the compound contains carbon in it).

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What is 0.00550 g converted to mg
Mila [183]

Answer: it is 5.5 mg

Explanation:

you have to multiply the mass value by 1000

5 0
3 years ago
CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

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

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

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

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
Inertia increases as an object's Blank Space __________ increases.
Anestetic [448]
<span>acceleration I think.</span>
4 0
3 years ago
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The lab just ran out of 1 M HCl that you need to complete the Benzillic Acid lab. The TA tells you there is 12 M HCl in the fume
Viktor [21]

Answer:

0.83 mL

Explanation:

Given data

  • Initial concentration (C₁): 12 M
  • Initial volume (V₁): ?
  • Final concentration (C₂): 1.0 M
  • Final volume (V₂): 10.0 mL

We can calculate the initial volume of HCl using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 1.0 M × 10.0 mL / 12 M

V₁ = 0.83 mL

The required volume of the initial solution is 0.83 mL.

7 0
3 years ago
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