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3 years ago

Which of the following allosterically activates mammalian isocitrate dehydrogenase?a. ADPb. NADHc. ATPd. MalateCitrate

Chemistry

1 answer:

Bogdan [553]3 years ago

5 0

Answer:

a. ADP

Explanation:

Isocitrate dehydrogenase is an enzyme in citric acid cycle that occurs in mitochondrial matrix.

It catalyzes conversion of the isocitrate to the alpha-ketoglutarate and Carbon dioxide which occurs in a two step process.

<u>IDH is regulated allosterically by the ADP positively in the mammals and inhibited by ATP, NADH.</u>

The enzyme, IDH will not catalyze the reaction unless there are low levels of ADP.

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Elements are organized into groups / families according to their physical and chemical properties. Identify the elements that's

Kryger [21]

Answer:

1 valence - Alkali Metals: Li Lithium, Na Sodium, K Potassium

2 valence - Alkaline Earth Metals: Be Beryllium, Mg Magnesium, Ca Calcium

3 valence - Non-metals: B Boron, Al Aluminium

4 valence - Non-metals: C Carbon, Si Silicon

5 valence - Non-metals: N Nitrogen, P Phosphorus

6 valence - Non-metals: O Oxygen, S Sulfur, Se Selenium

7 valence - Halogens: F Fluorine, Cl Chlorine, Br Bromine

8 valence - Noble Gases: He Helium, Ne Neon, Ar Argon

3 0

3 years ago

Cuántos electrones de Valencia tiene Ga

Zinaida [17]

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1

Primero realizas la configuración electrónica que es la que te puse allá arriba.

Después miras el nivel en que termina, puede ser 1, 2, 3, 4 etc.

Entonces como el último número de la configuración electrónica es 4, entonces ese es el nivel

Y los electrones de el último nivel son los de Valencia

4s2, 4p1 sumas 2+1 que son los electrones que se encuentran en el último nivel.

por eso hay 3 electrones de valencia.

4 0

3 years ago

1. When frequency increases energy:

liraira [26]

B

When frequency increases, as does the energy, but wavelength decreases. It also works vise versa; if wavelength were to increase, its frequency and energy will decrease.

8 0

3 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

3 years ago

Draw all four products obtained when 2-ethyl-3-methyl-1,3-cyclohexadiene is treated with HBr at room temperature and show the me

LenKa [72]

Answer:

See explanation below

Explanation:

In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.

At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.

Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.

In the picture below, you have the mechanism for all the 4 products.

Hope this helps

5 0

3 years ago