The hybridization state of the central sulfur atom is sp3d.
The valence shell electron pair repulsion theory describes the shape of molecules in terms of the number of electron pairs that surround the valence shell of the central atom in the molecule.
There are certain molecular shapes that are associated with certain hybridization states of the central atom. Since the central sulfur atom is trigonal bipyramidal, the hybridization state is sp3d.
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Answer:
a.all four carbon atoms.
Explanation:
Acetyl-CoA labeled with 14C in both of its acetate carbon atoms is incubated with unlabeled oxaloacetate and a crude tissue preparation capable of carrying out the reactions of the citric acid cycle. After one turn of the TCA cycle, oxaloacetate would have 14C in all four carbon atoms.
Answer: 8855 Joules energy is released when 115 g of 
changes from a liquid to a solid at its freezing point.
Explanation:
Latent heat of freezing is the amount of heat released to convert 1 mole of liquid to solid at atmospheric pressure.
Amount of heat released to freeze 1 gram of = 77 J/g
Mass of given = 115 gram
Heat released when 1 g of is freezed = 77 J
Thus Heat released when 115 g of is freezed =
Thus 8855 Joules energy is released when 115 g of changes from a liquid to a solid at its freezing point.
Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
