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zzz [600]
2 years ago
6

Modify the given molecule to show the product of the oxidation reaction using as the oxidizing agent. Include the appropriate hy

drogen atoms and charges. The molecule minus O O C C H 2 C H (O H) C O O minus is oxidized by N A D plus. Modify the given molecule to show the product of the oxidation reaction using as the oxidizing agent. Include the appropriate hydrogen atoms and charges. The molecule minus O O C C H 2 C H 2 C O O minus is oxidized by F A D plus.

Chemistry
1 answer:
Otrada [13]2 years ago
7 0

Answer:

NADH and FADH

Explanation:

An oxidizing agent is a chemical substance which has one among the three listed properties.

1. It can accept electrons.

2. It can remove hydrogen.

3. It can add oxygen to the reactant.

NAD+ and FAD+ serve as an oxidizing agent because they not only accept a pair of electron but also remove hydrogen. In this molecule, they have actually taken hydrogen from carbon where OH functional group was present and oxidizing that particular alcohol group to ketone. Please see the image attached.

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0.0010 micrometer = 0.001 x 10^-6 meter = 10^-9 meter = 0.00000001 meter
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2 years ago
From the following list of elements, those that will always form ionic compounds in a 1:2 ratio with zinc.
harina [27]

Answer:

A. iodine

C. fluorine

F. bromine

Explanation:

Ionic bonds occur mostly between metals and non-metals. Usually, a wide electronegativity difference is preferred between the two atoms. This makes one atom more desirous to gain electron and other more willing to donate electrons.

To have Zn forming  a compound in the ratio of 1 to 2, the combining power must be similar to this.

The dominant oxidation state of Zn is the is the +2 state.

The other combining atoms must have the ability to recieve the two electrons.

The halogens fit perfectly into this picture. They need just an electron to attain nobility. They are also highly electronegative. If two halogens combines with the Zn, then the ionic bond will result.

The halogens are fluorine,chlorine, bromine, iodine and astatine.

They will form these compounds:

                       ZnF₂, ZnBr₂ and ZnI₂

3 0
2 years ago
What volume of air contains 10.0g of oxygen gas at 273 k and 1.00 atm?
VARVARA [1.3K]
12.3g 02times1 mol 02/32.0 g 02=mol 02
.38mol/x=20.95/100
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7 0
2 years ago
Read 2 more answers
Which of the following species is not formed through a termination reaction in the chlorination of methane? Which of the followi
krek1111 [17]

Explanation:

When we add chlorine to a substance or compound then this process is known as chlorination.

For example, a process of chlorination is as follows.

Initiation : Cl_{2} \overset{light}{\rightarrow} 2Cl*          

where, Cl* is a free radical.

Propagation:

    Cl* + CH_{4} \rightarrow HCl + *CH_{3}

    CH_{3} + Cl_{2} \rightarrow CH_{3}Cl + Cl*

Termination:

    2Cl* \rightarrow Cl_{2}

    Cl* + *CH_{3} \rightarrow CH_{3}Cl

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Thus, we can conclude that out of the given options H_{2} is not formed through a termination reaction in the chlorination of methane.

5 0
2 years ago
What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter
Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, 31.0^oC

Explanation :

First we have to calculate the heat produced.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of NaOH = 1.52 g

Molar mass of NaOH = 40 g/mol

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole

Now put all the given values in the above formula, we get:

44.5kJ/mol=\frac{q}{0.038mol}

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Now we have to calculate the final temperature of solution in the calorimeter.

q=m\times c\times (T_2-T_1)

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = 4.18J/g^oC

T_1 = initial temperature = 20.1^oC

T_2 = final temperature = ?

Now put all the given values in the above formula, we get:

1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)

T_2=31.0^oC

Thus, the final temperature of the solution in the calorimeter is, 31.0^oC

4 0
2 years ago
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