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2 years ago

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$\mathcal\blue{what \: is \: organic \: chemistry}$
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Chemistry

2 answers:

nignag [31]2 years ago

6 0

Organic chemistry as the the study of general properties and compositions of organic compounds.

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<h3>What is organic chemistry?</h3>

Organic chemistry can be simply defined as the study of organic compounds.

Organic chemistry studies the structure, properties, composition, reactions, and preparation of carbon-containing compounds also known as organic compounds.

Thus, we can defined organic chemistry as the the study of general properties and compositions of organic compounds.

Learn more about organic chemistry here: brainly.com/question/704297

svlad2 [7]2 years ago

6 0

Organic chemistry is the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds

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The following pieces of evidence were found at separate explosion sites. For each item, indicate whether the explosion was more

tangare [24]

Explanation:

The major difference between low and high explosives is the rate of detonation. Low explosives detonate very slowly (less than 1,000 meters per second), whereas high explosives detonate very quickly (from 1,000 to 8,500 meters per second).

High explosives among the given list are Lead azide residues, Ammonium nitrate residues, and Scraps of primacord. Whereas Nitrocellulose residues and, Potassium chlorate residues are low explosives.

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3 years ago

If the temperature of an area drops 8 degrees from one day the next the climate has changed.

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3 years ago

A balloon filled with 1.92 g of helium has a volume of 12.5 L. What is the balloon’s volume after 0.850 g of helium has leaked o

Marizza181 [45]

The answer is 1.56L. Avogadro's Law states that the volume of a gas is directly proportional to the number of moles (or a number of particles) of gas when the temperature and pressure are held constant.

V∝n

V₁/n₁m= V₂/n₂

V₁ = initial volume of gas = 12.5 L

V₂ = final volume of gas = ?

n₁ = initial moles of gas = 0.016 mole

n₂ = final moles of gas = 0.016-0.007 = 0.002 mole

V₁/n₁m= V₂/n₂

V₂= 1.56L

Avogadro's Law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up.

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8 0

1 year ago

Write an equation that shows the formation of the chloride ion from a neutral chlorine atom

jeka57 [31]

The gaining of electron by an atom results in the formation of anion shown by the negative charge on the atom whereas lose of electron results in the formation of cation shown by positive charge on the atom. The atom lose or gain electron to complete their octet and get stable in nature.

The chlorine atom will gain an electron and form chloride anion with one negative charge on it. The chloride ion is more stable in nature compared to the chlorine atom due to complete octet of chloride ion by gaining of electron.

Electronic configuration of chlorine atom is:

$[Ne]3s^{2}3p^{5}$

By gaining of one electron, electronic configuration of chloride ion is:

$[Ne]3s^{2}3p^{6}$

Thus, the equation that shows the formation of the chloride ion from a neutral chlorine atom is:

$Cl+1e^{-}\rightarrow Cl^{-}$

4 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago