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Annette [7]
3 years ago
10

On the basis of the information above, a buffer with a pH = 9 can best be made by using

Chemistry
1 answer:
telo118 [61]3 years ago
3 0

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for \mathbf{H_3PO_4 , H_2PO^{-}_4 ,  HPO_4^{2-}} are \mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \  5\times 10^{-13}} respectively.

\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}

\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}

\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}

The reason while option D is the best answer is that, the value of pKa for both

\mathbf{H_2PO^{-}_4 ,\  \& \  HPO_4^{2-}} lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}

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bulgar [2K]

Answer:

likely be the same

Explanation:

this is because we have one color that both atoms share (green). both sample 1 and sample 2 have green and another color. yet, since they share one color, they are likely similar

7 0
3 years ago
Which element is reduced in this reaction? 2kmno4 3na2so3 h2o→2mno2 3na2so4 2koh
erik [133]

In a redox chemical reaction, one species gets reduced and another gets oxidized. Manganese element is reduced in this reaction.

<h3>What is oxidized and reduced?</h3>

In a redox reaction, the increase or decrease in the oxidation number and electrons results in the reduction and oxidation of the chemical species. The oxidation and reduction occur simultaneously in a reaction.

The oxidation number of Mn in permanganate ion was +8 on the left side and decreased to +4 on the right side of the equation. Potassium permanganate is an oxidizing agent that has reduced the manganese ion of the permanganate ion.

Therefore, manganese is reduced.

Learn more about reduction and oxidation here:

brainly.com/question/2427143

#SPJ4

3 0
2 years ago
. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are
olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

K_{sp}\text{ of }Ag_2S=8\times 10^{-51}

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K

where,

\Delta G^o = standard Gibbs free energy = ?

R = Gas constant = 8.314J/K mol

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

K = equilibrium constant or solubility product = 8\times 10^{-51}

Putting values in above equation, we get:

\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ

For the given chemical equation:

Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)

The equation used to calculate Gibbs free change is of a reaction is:  

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

The equation for the Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]

We are given:

\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ

Putting values in above equation, we get:

285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol

Hence, the standard free energy change of formation of S^{2-}(aq.) is 92.094 kJ/mol

8 0
3 years ago
Help with Organic chemistry mechanism
yarga [219]
3H + 3Br = HBr9 Organic chemistry mechanism
4 0
3 years ago
Given each pair, complete the sentences to determine which member of each has the stronger intermolecular dispersion forces.
evablogger [386]

Answer:

CH3CH2CH2Cl

CH3CH2CH2CH2CH2SH

Br2

Explanation:

Dispersion forces increases with increase in relative molecular mass. The specie having the greater relative molecular mass definitely has greater dispersion forces. A rough estimation of the relative molecular masses of the species stated in the answer will reveal this fact.

3 0
3 years ago
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