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4 years ago

2 bromocyclopentanamine structural formula

Chemistry

1 answer:

Elis [28]4 years ago

5 0

Answer:

The structure is given in attached file.

Explanation:

Explanation

2-bromocyclopentamine (Figure attached) is a synthetic compound which is synthesized by substitution reaction of cyclopentamine and hydrobromide. Its molecular formula and molecular mass are C5H10NBr and 164.05 mol/g respectively. It is a very reactive compound so it doesn’t available in pure form, it is present in market as a mixture of 2-bromocyclopentamine and Hydrobromide.

Properties :

Its boiling point is 115 0C

Its melting point is – 75 oC

It is highly flammable

It is highly toxic

It is irritant

It is corrosive in nature

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Why elements in the same group have the same charge

frozen [14]

In many cases, elements that belong to the same group (vertical column) on the periodic table form ions with the same charge because they have the same number of valence electrons.

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3 years ago

is carried out adiabatically in a constant-volume batch reactor. The rate law is Plot and analyze the conversion, temperature, a

Sphinxa [80]

Answer:

See explaination

Explanation:

The mole balance for a constant-volume batch reactor is given such as, For a first-order isothermal reaction, the time to reach a given conversion is the same for constant-pressure and constant-volume reactors. Also, the time is the same for a reaction of any order if there is no change in the number of moles.

Please kindly check attachment for the step by step solution of the given problem.

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3 years ago

What concentration of OH is required to created a pOH of 1.3.

Crank

Answer:

0.0501

Explanation:

OH=10^-poH

OH=10^-1.3

=0.0501

6 0

3 years ago

The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00

Marina CMI [18]

<u>Answer:</u> The moles of $CO_2$ added to the system is 7.13 moles

<u>Explanation:</u>

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

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3 years ago

Where do the numbers in a stoichiometry mole ratio come from?

erik [133]

it is defined as the ratio of moles of one substance to the moles of another substance in a balanced equation. To determine the mole ratio between two substances, all you need to do is look at the balanced equation for the coefficients in front of the substances you are interested in.

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3 years ago