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3 years ago

7

Draw the beta-hydroxy Ketone Intermediate and final condensation product for the following Aldol Reaction: Yol 2200 1 equiv. 2 e

quiv

1 answer:

jeka943 years ago

6 0

Answer:

idktdhfff jxjf

Explanation:

nopefvffgh gfgg

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Cerium (IV) ions are strong oxidizing agents in acid ic solution, oxidizing arsenio us acid to arsen ic acid accor ding to the f

kirill [66]

Answer:

0.2042 M is the original concentration of $Ce^{4+}$ (aq) in the titrating solution.

Explanation:

Mass of $As_2O_3$ = 0.217 g

Moles of $As_2O_3=\frac{0.217 g}{198 g/mol}=0.001096 mol$

1 mole of $As_2O_3$ have 2 mole of As and 1 mole of $H_3AsO_3$ have 1 mole of As.

So, from 1 mole of $As_2O_3$ we will have 2 moles of $H_3AsO_3$

Then from 0.001096 mol of $As_2O_3$ :

$2\times 0.001096 mol=0.002192 mol$ of $H_3AsO_3$

$2Ce^{4+}(aq)+H_3AsO_3(aq)+3H_2O(l)\rightarrow 2Ce^{3+}(aq)+H_3AsO_4(aq)+2H^+(aq)$

According to reaction, 1 mole of $H_3AsO_3$ reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of

$\frac{2}{1}\times 0.002192 mol=0.004384 mol$ of cerium (IV) ions.

Volume of the acidic cerium{IV) sulfate = 21.47 ml =0.02147 L

1 mL = 0.001 L

$concentration = \frac{Moles}{Volume(L)}$

$[Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M$

0.2042 M is the original concentration of $Ce^{4+}$ (aq) in the titrating solution.

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3 years ago

118g of o2 gas is held at STP what is the volume of gas

shutvik [7]

Answer:

At STP one mole of any gas occupies a volume of 22.4 L: this is the molar volume.

Explanation:

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4 years ago

Which gas law refers to the solubility of a gas changing with the pressure over the solution

Dmitry [639]

Answer:

the answer should be henry's law

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4 years ago

On which sides of the Earth do high tides occur?

soldi70 [24.7K]

Tides move around Earth as bulges in the ocean. As the ocean bulges toward the moon, a high tide is created. The high tide on the side of Earth facing the moon is called the high high tide. The high tide caused by the bulge on the opposite side of Earth is called the low high tide.

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3 years ago

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces

Alchen [17]

The question is incomplete, here is the complete question:

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 5.0 L flask with 2.2 atm of ammonia gas and 2.4 atm of oxygen gas at 44.0°C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.99 atm.

Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

<u>Answer:</u> The pressure equilibrium constant for the reaction is 32908.46

<u>Explanation:</u>

We are given

Initial partial pressure of ammonia = 2.2 atm

Initial partial pressure of oxygen gas = 2.4 atm

Equilibrium partial pressure of nitrogen gas = 0.99 atm

The chemical equation for the reaction of ammonia and oxygen gas follows:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

<u>Initial:</u> 2.2 2.4

<u>At eqllm:</u> 2.2-4x 2.4-3x 2x 6x

Evaluating the value of 'x':

$\Rightarrow 2x=0.99\\\\x=0.495$

So, equilibrium partial pressure of ammonia = (2.2 - 4x) = [2.2 - 4(0.495)] = 0.22 atm

Equilibrium partial pressure of oxygen gas = (2.4 - 3x) = [2.4 - 3(0.495)] = 0.915 atm

Equilibrium partial pressure of water vapor = 6x = (6 × 0.495) = 1.98 atm

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{N_2})^2\times (p_{H_2O})^6}{(p_{NH_3})^4\times (p_{O_2})^3}$

Putting values in above equation, we get:

$K_p=\frac{(0.99)^2\times (1.98)^6}{(0.22)^4\times (0.915)^3}\\\\K_p=32908.46$

Hence, the pressure equilibrium constant for the reaction is 32908.46

5 0

3 years ago