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3 years ago

Which statements describe inorganic compounds? Check all that apply.

1 answer:

6 0

Inorganic compounds contain carbon. Inorganic compounds usually lack carbon. Inorganic compounds are not associated with or made from living things

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Which of the following is the electrical charge on an atom that contains 6 electrons, 3 neutrons, and 4 protons?

Y_Kistochka [10]

Answer:

Its electrical charge is -2.

4 0

3 years ago

In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit

Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

(1) Preparatory phase

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.

(2) Pay-off phase

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that BPG has a mixed anhydride and the bond at C1 is a very high energy bond. In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

3 0

3 years ago

PLEASE HELP ME!!! It would be great

kompoz [17]

The answer should be 0.5 because 6/12 is 0.5. It gives u the key equation

3 0

3 years ago

3/4(x-2)=x-1 please answer me

Elan Coil [88]

Answer:

-2

Explanation:

3/2 (x-2) = x-1

<=> 3/4 x - 3/2 = x-1

<=> 3/4 x - x = 3/2 -1

<=> -1/4 x =1/2

<=> x= -2

3 0

3 years ago

Read 2 more answers

A plot of binding energy per nucleon (Eb/ A) versus the mass number (A) shows that nuclei with a small mass number have a small

juin [17]

Answer:

a) 1.12 MeV / nucleon

b) 5.62 MeV / nucleon

c) 8.80 MeV / nucleon

d) 8.56 MeV / nucleon

we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

Explanation:

Binding energy = ( Δm * 931.5 ) MeV

Binding energy per nucleon = Binding energy in / Number of nucleon

a) ²H = 1 neutron , 1 proton = 2 nucleons

Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

∴ Binding energy per nucleon = ( 2.4 * 10^-3 * 931.5 ) MeV / 2 nucleons

= 1.12 MeV / nucleon

b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons

theoretical mass = 7.0160 u

Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 ) = 7.0582 u

Δm = ( 7.0582 u - 7.0160 u ) = 0.0422 u

∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7

= 5.62 MeV / nucleon

C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons

Theoretical mass = 61.9283 u

Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

= 62.5142 u

Δm = 0.5859 u

∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62

= 8.80 MeV / nucleon

D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons

Theoretical mass = 109.9030 u

Actual mass = ( 48 * 1.0078 ) + ( 62 * 1.0087 )

= 110.9138 u

Δm = ( 110.9138 - 109.9030 ) = 1.0108 u

∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110

= 8.56 MeV / nucleon

hence we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

3 0

3 years ago