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Anna11 [10]
2 years ago
15

A compound with the molecular formula C10H10O4 produces a 1H NMR spectrum that exhibits only two signals, both singlets. One sig

nal appears at 3.9 ppm with a relative integration value of 79. The other signal appears at 8.1 ppm with a relative integration value of 52. Draw the structure of this compound.

Chemistry
1 answer:
Kamila [148]2 years ago
8 0

Answer:

The attached figure shows the structure of dimethyl terephthalate.

Explanation:

Dimethyl terephthalate is a compound whose formula is C6H4 (COOCH3) 2. It is a diester produced from terephthalic acid and methanol. It is characterized by being a white solid. Another method for the preparation is from p-xylene and methanol, which is characterized by having an oxidation and an esterification.

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eleanor purchased $2568 worth of stock and paid her broker a 0.5% fee. She sold the stock when the stock price increased to 3928
Mrrafil [7]

Answer: $1338.16

Explanation: Total cost of stock= $2568

Total cost of stock including the brokerage =2568+\frac{0.5}{100}\times {2568}=2580.84$

Selling price of stock = $3928

Selling price of stock including trading fee=($3928-$7)=$3919

Net Proceeds = Net selling price of stock - Cost Price of stock

Net Proceeds = ($3919 - $2580.84) = $1338.16


5 0
2 years ago
In the redox conversion of Ni2+ to NiO4−, the oxidation number of Ni goes from (−2, 0, +2) to (−1, +1, +7, +9). Recall that the
inessss [21]
Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero, 
for NiO4-, assuming oxidation number of Ni is x 
 (x ×1) + (-2 × 4) = -1
  x = + 7
Therefore, the oxidation number goes from 0 to +7
7 0
2 years ago
Calculate the pH of each of following buffered solutions.?a. 0.10 M acetic acid/0.25 M sodium acetate b. 0.25 M acetic acid/0.10
Masteriza [31]

Answer:

a. 5.10.

b. 4.35.

c. 5.10.

d. 4.35.

Explanation:

<u><em>a. 0.10 M acetic acid/0.25 M sodium acetate </em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.25)/(0.10)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

<u><em>b. 0.25 M acetic acid/0.10 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.10)/(0.25)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

<u><em>c. 0.080 M acetic acid/0.20 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.20)/(0.08)

∴ pH =  4.744 + 0.34 = 5.084 ≅ 5.10.

<u><em></em></u>

<u><em>d. 0.20 M acetic acid/0.080 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.08)/(0.20)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

6 0
2 years ago
What is the poh of an aqueous solution at 25.0 °c that contains 1.35 × 10-8 m hydroxide ion?
ohaa [14]
POH is defined as the negative log base ten of the concentration of base. You have the concentration.
8 0
2 years ago
Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −
Grace [21]

Answer : The value of K_{eq} is, 11.2

The value of \Delta G_{rxn} is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

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

K_{eq}  = equilibrium constant  = ?

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

\Delta G^o=-RT\times \ln K_{eq}

-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}

K_{eq}=11.2

Thus, the value of K_{eq} is, 11.2

Now we have to calculate the \Delta G_{rxn}.

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

The given reaction is:

A(aq)\rightleftharpoons B(aq)

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

\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}    ............(1)

where,

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

\Delta G_^o =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol

Therefore, the value of \Delta G_{rxn} is -9.04 kJ/mol

3 0
2 years ago
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