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3 years ago

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Please help :) Explain why a change in the position of a substituent group can, but does not always, result in a different compo

und. Give an example of when changing the position of this group results in a different compound and another example when it does not result in a different compound.

1 answer:

Lady bird [3.3K]3 years ago

6 0

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Which feature of a rosebush is an adaptation for defense?

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sharp thorns

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What are the parts of an atom, and how are they used to determine an atom’s mass and identity?

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4 years ago

The name given to the main group elements in group 2 is _______.

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3 years ago

1. A solution is made by mixing 50mL of 2.0M K2HPO4 and 25mL of 2.0M KH2PO4. The solution is diluted to a final volume of 200mL.

Kazeer [188]

Answer:

The pH of the final solution is 7.15

Explanation:

50 mL of 2.0 M of $K_2HPO_4$ and 25 mL of 2.0 M of $KH_2PO_4$ were mixed to make a solution

Final volume of the solution after dilution = 200 mL

Final concentration of $K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M$

Final concentration of $KH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M$

We use Hasselbach- Henderson equation:

$pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85$

Substituting the values:

$pH = 6.85+ log \frac{0.5}{0.25}pH = 6.85+ log 2pH = 6.85+ 0.3 = 7.15$

Therfore, the pH of the final solution is 7.15

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3 years ago

1. (a) Write the two electrochemical half reactions for ethanol (C2H5OH) direct electrochemical conversion in a fuel cell with a

Natalka [10]

Answer:

(a) Two electrochemical half reactions for ethanol $(C_{2}H_{5}OH)$ is written below

$C_{2}H_{5} + 60$ ⇒ $2CO_{2} + 3H_{2}O + 12e^{-} (anode)$

$3O_{2} + 12e^{-}$ ⇒ $60^{=}$ $(Cathode)$

(b) The direct oxidation of the fuel will occur in a solid oxide fuel cell but we have to compete with other sets of chemical electrochemical reaction such as water-gas-shift reaction

$CO + H_{2}O$ ⇆ $CO_{2} + H_{2}(WGSR)$

(c) The electrochemical half reaction that could convert ethanol directly into a fuel cell that conducts hydrogen ions is shown below

$C_{2}H_{5}OH + 3H_{2}O$ ⇒ $2CO_{2} + 12H^{+} + 12e^{-} (anode)$

$3O_{2} + 12H^{+} + 12e^{-} - 6H_{2}O (Cathode)$

$C_{2} H_{5} OH + 3O_{2}$ ⇒ $3H_{2}O + 2CO_{2}$

(d) The half reactions that would be required are shown below

$C_{2}H_{5} OH + 12OH^{-}$ ⇒ $CO_{2} + 9H_{2}O + 12e^{-} (anode)$

$3O_{2} + 6H_{2}O + 12e^{-}$ ⇒ $12OH^{-} (Cathode)$

$C_{2}H_{5}OH + 3O_{2}$ ⇒ $3H_{2}O + 2CO_{2}$

8 0

4 years ago