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Free_Kalibri [48]
2 years ago
9

Assume the following potential energies for a Newman projection: anti Me/H or anti Me/Me = 0 kcal/mole; eclipsed H/H = 4 kcal/mo

le; eclipsed Me/Me = 8 kcal/mole; gauche H/H and gauche H/Me = 0 kcal/mole; gauche Me/Me = 4 kcal/mole For 2,3-dimethylbutane, when drawing a Newman projection viewing down the C2-C3 bond, what is the potential energy of the most stable Newman projection?
Chemistry
1 answer:
kap26 [50]2 years ago
4 0

Answer:

0 kcal/mole

Explanation:

The most stable Newman projection corresponds to the minimum potential energy: Answer (e) 0 kcal/mole

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What is the ph of a solution of 0.400 m k2hpo4, potassium hydrogen phosphate?
dusya [7]
When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :

when Pka = - ㏒ Ka

          6.86 = -㏒ Ka 

∴Ka = 1.38 x 10^-7

by using ICE table:

               H2PO4- →  H+  + HPO4
initial      0.4 m            0         0

change     -X                +X       +X

Equ       (0.4-X)               X          X

when Ka = [H+][HPO4] / [H2PO4-]

by substitution:

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∴PH = -㏒[H+]

        = -㏒ (2.3 x 10^-4)
 ∴PH  =  3.6

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2 years ago
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