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3 years ago

Write the balanced oxidation-reduction reaction equation for the oxidation of benzoin by ammonium nitrate. 2

1 answer:

7 0

Benzoin is an organic compound with the formula PhCH(OH)C(O)Ph. It is a hydroxy ketone attached to two phenyl groups.

To answer your question, the balanced oxidation-reduction reaction equation for the oxidation of benzoin by ammonium nitrate is:

2Ph-C(OH)-C(O)-Ph+NH4NO3 --> 2Ph-C(O)-C(O)-Ph + N2 + 3H2O.

I hope this helps and if you have any further questions, please don’t hesitate to ask again.

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Which of the following events takes place in the Kreb entry phase (acetyl COA from pyruvate)?

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Answer:

Alphabet C :NADH is oxidized,CO2 is reduced

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3 years ago

List several other words that are formed using the prefix mon or mono what do these words have in common

UkoKoshka [18]

Answer:

1) Monochrome

2) Monocot

3) Monosaccharide

4) Monounsaturated

5) Monoxide

Explanation:

The prefix mono or mon is of Greek origin and it means single, lone or one.

From the meaning of the words we can tell that they have something in common:

1) Monochrome: This refers to an artwork, a painting or drawing that has just one color.

2) Monocot: This is a type of flowering plant that has a single embryo and a single cotyledon.

3) Monosaccharide: This is a type of simple sugar that contains one sugar unit that cannot be split.

4) Monounsaturated: This refers to a molecule that contains one double or triple bond.

5) Monoxide: This refers to a form of oxide that contains only one oxygen atom.

From the definitions above, all the words mean having only one or a single item.

That is what they have in common.

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3 years ago

Ethylene glycol (C2H6O2) is used as an antifreeze in cars. If 450 g of ethylene glycol is added to 4.25 kg of water, what is the

slamgirl [31]

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3 0

3 years ago

Gaseous indium dihydride is formed from the elements at elevated temperature:

3241004551 [841]

Answer:

For 1: The value of $Q_p$ for above reaction is 36.83

For 2: The value of $Q_p$ for above reaction is 36.83

For 3: The equilibrium partial pressure of Indium is 0.126 atm

For 4: The equilibrium partial pressure of hydrogen gas is 0.094 atm

For 5: The equilibrium partial pressure of Indium dihydrogen is 0.018 atm

Explanation:

We are given:

Partial pressure of Indium gas = 0.0650 atm

Partial pressure of hydrogen gas = 0.0330 atm

Partial pressure of Indium dihydride = 0.0790 atm

The given chemical equation follows:

$ln(g)+H_2(g)\rightleftharpoons InH_2(g)$

Initial: 0.065 0.033 0.079

At eqllm: 0.065-x 0.033-x 0.079+x

For 1:

The expression of $Q_p$ for above reaction follows:

$Q_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$Q_p=\frac{0.079}{0.065\times 0.033}=36.83$

Hence, the value of $Q_p$ for above reaction is 36.83

For 2:

We are given:

$K_p$ of the reaction = 1.48

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

As, $Q_{p}>K_p$ for the given reaction, the reaction is reactant favored.

Hence, the reaction proceed in the backward direction to attain equilibrium

For 3:

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$1.48=\frac{(0.079+x)}{(0.065-x)\times (0.033-x)}\\\\x=-0.061,0.835$

Neglecting the value of x = 0.835 because the reaction is going backwards. So, by taking this value, the pressure of the reactants will decrease

So, equilibrium partial pressure of Indium = (0.065 - x) = [0.065 - (-0.061)] = 0.126 atm

For 4:

The equilibrium partial pressure of hydrogen gas = (0.033 - x) = [0.033 - (-0.061)] = 0.094 atm

For 5:

The equilibrium partial pressure of Indium dihydrogen = (0.079 + x) = [0.079 + (-0.061)] = 0.018 atm

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3 years ago

How can an acid salt be converted to a normal salt

erik [133]

Neutralize it with a alkaline

7 0

4 years ago