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Ksenya-84 [330]
3 years ago
6

Calculate the value deltaG°​

Chemistry
1 answer:
atroni [7]3 years ago
4 0

Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

Explanation:

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ahrayia [7]

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6 0
3 years ago
Nitric acid (HNO3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.
Alborosie

<u>Given:</u>

Concentration of HNO3 = 7.50 M

% dissociation of HNO3 = 33%

<u>To determine:</u>

The Ka of HNO3

<u>Explanation:</u>

Based on the given data

[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M

The dissociation equilibrium is-

            HNO3   ↔    H+      +      NO3-

I            7.50               0                 0

C          -2.48          +2.48              +2.48

E            5.02            2.48              2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

6 0
3 years ago
Which statement below is not true?
Verizon [17]
Hi I believe the answer is B
Hope this helps you
7 0
2 years ago
The essential oil found in cloves, eugenol, can be isolated by steam distillation because it is insoluble in water and has a mea
Damm [24]
We know that:
Molar Mass H2O: 18 g/mol 
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>

<span>Using formula:
V= [mole fraction x molar mass] / density </span>

<span>mH20: 0.9947 * 18
          = 17.9046 / 1 g/mL
          = 17.9046 </span>
<span>morg: 0.0053 * 164  
        = 0.8692/ 1.05 g/mL
        = 0.8278 </span>

<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>

Yotal volume = 30 mL; therefore, 
<span>0.0442 = (volume eugenol/30) </span>

<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
            = 1.44/1.05
            = about 1.37 mL </span>
6 0
2 years ago
A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
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