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steposvetlana [31]
2 years ago
8

4. If 0.130 g of p-aminophenol is allowed to react with excess acetic anhydride, what is the theoretical yield of acetaminophen

in moles? In grams?
Chemistry
1 answer:
Dominik [7]2 years ago
8 0

Hey There!

p-aminophenol (109.13 g/mol) + Ac2O (102.09 g/mol) ---> acetominophen (151.16 g/mol)

next, since you already know your limiting reactant (p-aminophenol), convert it to mols  :

0.130 g / (109.13 g/mol) = 0.00119 moles

now that's your theoretical max, since its a 1:1 mol ratio, so multiply by the new molecular weight. :

0.00119  * 151.16  = 0.180 g

Hope that helps!

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Calculate the temperature in k of 3.05 moles of gas occupying 3.70 L at 4.12 atm
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Answer:

{ \bf{PV = nRT}} \\ { \tt{(4.12 \times 3700) = 3.05 \times 0.083 \times T }} \\ { \tt{15244 = 0.25315 \: T}} \\ { \tt{T = 6.02 \times  {10}^{4} \: kelvin }}

6 0
2 years ago
For the following SN2 reaction, draw the organic and inorganic products of the reaction, and identify the nucleophile, substrate
Sauron [17]

Answer:

Substrate:alkyl halide

Leaving group: Cl

Organic product: The nitrile

Inorganic product: Cl-

Nucleophile: CN-

Explanation:

An SN2 reaction is a concerted bimolecular reaction. Concerted means that it involves two reactions taking place at the same time while bimolecular means that the rate determining step involves two molecules. The cyanide ion attacks the alkyl halide from the rear. In the transition state, the leaving group (Cl-) is departing while the nucleophile (CN-) is forming a bond to the alkyl halide simultaneously. The alkyl halide is the substrate in the reaction. The organic product is the nitrile shown in the image attached.

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What are the Products in this Chemical Reaction? PLEASE ANSWER ASAP!!
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2 years ago
How many molecular orbitals are present in the valence band of a sodium crystal with a mass of 4.14 g
Neko [114]

Answer:

Explanation:

Given that:

The mass of sodium crystal = 4.14 g

As a result, the number of atoms in the sodium crystal is:

=4.14 \g \times \dfrac{ 1 \ mol}{23.0 \ g }\times \dfrac{6.023 \times 10^{23} \ atoms}{1 \ mol}

= 1.084 × 10²³ atoms

In the valence orbital of the sodium atom, there is just one electron.

As a result, the number of molecular orbitals produced by these atoms is= 1.084 × 10²³ molecular orbital

Because valence bands are created from occupied molecular orbitals, the number of valence bands created from the aforementioned molecular orbitals is:

= \dfrac{ 1.084 \times 10^{23}}{2}

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6 0
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Q.11. Calculate the mass of NaBr needed to prepare a 50.00-mL aqueous solution that would yield the same conductivity as the NaC
klemol [59]

Answer:

The mass of NaBr needed is 0.22969 g.

Explanation:

1 mole of NaBr contains 22.4 dm^3 of NaBr

Therefore, 0.05 dm^3 (50/1000 = 0.05 dm^3) of NaBr = 0.05/22.4 = 0.00223 mol of NaBr

MW of NaBr = 23 + 80 = 103 g/mol

Mass of NaBr = number of moles × MW = 0.00223 × 103 = 0.22969 g

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2 years ago
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