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ICE Princess25 [194]
4 years ago
15

Why do all the elements in Group 16 share similar properties?

Chemistry
1 answer:
geniusboy [140]4 years ago
5 0
D. They all have the same number of valence electrons.
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Can someone please help me out?
ziro4ka [17]
Yes the answer is c because the atoms are preserved
4 0
3 years ago
Read 2 more answers
Fructose‑2,6‑bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolys
olga_2 [115]

Answer:

Activate glycolysis/Inhibit gluconeogenesis: Increased levels of fructose-2,6-bisphosphate, activation of PFK-2

Activate gluconeogenesis/Inhibit glycolysis: Increased levels of glucagon, Increased levels of cAMP, Activation of fructose-2,6-bisphosphatase (FBPase-2)

<em>Note: The question is incomplete. The complete question is given below and in the attachment.</em>

<em>Fructose‑2,6‑bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose‑1,6‑bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose‑2,6‑bisphosphate is regulated by many hormones, second messengers, and enzymes. How do the following affect glycolysis and gluconeogenesis? </em>

Explanation:

Fructose-2,6-bisphosphate is an allosteric effector for the enzymes phosphofructokinase-1 (PFK-1) and fructose-1,6-bisphosphatase (FBPase-1). It increases the affinity of PFK-1 for fructose-6-phosphate thereby activating glycolysis. However, it reduces the affinity of FBPase-1 for its substrate, fructose-1,6-bisphosphate thereby inhibiting gluconeogenesis.

Activation of phosphofructokinase-2 activates glycolysis and inhibits gluconeogenesis by catalyzing the phosphorylation of fructose-6-phosphate to form fructose-2,6-bisphosphate.

Increased levels of glucagon stimulates the synthesis of cAMpP which activates cAMP-dependent ptrotein kinase which phosphorylates the bifunctional enzyme PFK-2/FBPase-2. The phosphorylation of this enzyme inhibits its PFK-2 activity and activates its FBPase-2 activity. This results in the activation of gluconeogenesis and inhibition of glycolysis.

Fructose-2,6-bisphosphatase breaks down fructose-2,6-bisphosphate to fructose-6-phospshate and a phosphoryl group. This results in the activation of gluconeogenesis and the inhibition of glycolysis.

7 0
4 years ago
Why sulphur has two melting point values
Fofino [41]

I don't know how well known/accepted this is (it's in my textbook so I'm guessing it's right), but Sulphur has two forms - the alpha and beta forms ,apparently gamma sulphur exists as well.

The alpha form is rhombic, yellow in color and has a MP of 385.8 K. The beta form is colorless and has a MP of 393 K and is formed by melting rhombic sulphur and cooling it till a crust forms on top. Poke a hole and pour out the liquid inside and you get beta sulphur. The transition point is 369K - below it, alpha sulphur is stable and above it, beta sulphur is stable. Both have helped. I had to pull out an old textbook and that's something that I don't usually do.

8 0
3 years ago
Which statement is true about energy changes?
Sindrei [870]

Answer:

D

Explanation:

energy can not be created or destroyed it can only be transferred or changed to one form to another.

7 0
4 years ago
You have a ballon filled with hydrogen gas which keeps it at a constant pressure, regardless the volume. The initial volume of t
abruzzese [7]

Answer:

619°C

Explanation:

Given data:

Initial volume of gas = 736 mL

Initial temperature = 15.0°C

Final volume of gas = 2.28 L

Final temperature = ?

Solution:

Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)

Initial temperature = 15.0°C (15+273 = 288 K)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 2.28 L × 288 K / 0.736 L

T₂ = 656.6 L.K / 0.736 L

T₂ = 892.2 K

K to °C:

892.2 - 273.15 = 619°C

7 0
3 years ago
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