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3 years ago

Which of the following classes of organic compounds has C=C in it?

1 answer:

4 0

The class of organic compounds that contain carbon-carbon double bonds, represented as C=C, is known as A) alkene.
Alkenes are unsaturated organic compounds and contain two carbon atoms bonded by one sigma bond and one pi bond. Alkenes exhibit different properties from alkanes due to the negative charge concentration at the double bond. Alkanes may be converted to alkenes and vice-versa.

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In what ways can geographic position be considered a temperature control?

Lesechka [4]

Near water, change in elevation, or change in latitude.

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2 years ago

What is the percent composition by mass of hydrogen in nh4hco3

Simora [160]

Answer: The percent composition by mass of hydrogen in given compound is 6.33 %

Explanation:

We are given:

A chemical compound having chemical formula of $NH_4HCO_3$

It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms

To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:

$\%\text{ composition of Hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of compound}}\times 100$

Mass of compound = $[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol$

Mass of hydrogen = $(5\times 1)=5g/mol$

Putting values in above equation, we get:

$\%\text{ composition of Hydrogen}=\frac{5g/mol}{79g/mol}\times 100=6.33\%$

Hence, the percent composition by mass of hydrogen in given compound is 6.33 %

6 0

3 years ago

What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlor

ddd [48]

Answer:

$1.75272\ \text{m}^3$

Explanation:

The breakdown reaction of ozone is as follows

$CF_3Cl + UV \rightarrow CF_3 + Cl$

$Cl + O_3 \rightarrow ClO + O_2$

$O_3 + UV \rightarrow O_2 + O$

$ClO + O \rightarrow Cl + O_2$

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

m = Mass of $CF_3Cl$ = 15.5 g

M = Molar mass of $CF_3Cl$ = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = $62.363\ \text{L mmHg/K mol}$

Number of moles is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}$

$20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}$

From ideal gas law we have

$PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3$

For 20 cycles of the reaction the volume of the ozone is $1.75272\ \text{m}^3$ .

7 0

3 years ago

Why is Avogadro’s number referred to as a mole? If you could change this terminology, what would you change it to and why?

stepan [7]

The number of particles (molecules, atoms, compounds, etc.) per mole of a substances is known as Avagadro number. It is equal to 6.022×10^23 mol-1 and is expressed as NA.

Number of moles is the amount of a substance that contains as many particles as there are atoms in 12 grams of pure carbon-12. So, 1 mol contains 6.022×10^23 elementary entities of the substance. Since 6.022 x 10^23 is the Avagadro number, one mole is equal to Avagadro number.

One mole of a substance is the ratio of mass of the substance by the molecular mass of the substance. Thus the mass of one mole of a substance is equal to the substance's molecular weight. Thus one mole of a substance is the atomic mass unit of a substance and since one mole is equivalent to the Avagadro number,we can conclude that one Avagadro number is one atomic mass unit of the substance.

7 0

3 years ago

In the laboratory you dissolve 15.9 g of barium chloride in a volumetric flask and add water to a total volume of 375 ml. what i

ASHA 777 [7]

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. So, we calculate as follows:

Molarity = 15.9 g BaCl2 ( 1 mol / 208.23 g ) / .375 L = 0.204 mol / L

5 0

3 years ago