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Veseljchak [2.6K]
3 years ago
9

How many traditional areas of study can chemistry be divided into

Chemistry
1 answer:
seraphim [82]3 years ago
6 0
<span>There is five main area of study in Chemistry, these are:

Analytical, this focusses on experimental equipment and methods used in chemistry (e.g., NMR, Spectroscopic methods, etc.)

Biochemistry - focuses on the chemistry of compounds and processes in living things (e.g., amino acids, proteins, DNA, cellular respiration, Krebs cycle, etc.)

Organic - focuses on the chemistry on most carbon-based molecules found in living things (e.g., hydrocarbons, alcohols, carbolic acids. Amines, ester, etc.)

Inorganic - (focuses on all elements other than carbon (e.g., fluorine, silicon, xenon, etc.)

Physical - focuses on the basic structure and energetic son atoms and molecules (e.g., subatomic structure, is nice and covalent bonding, thermodynamics, reactions, etc.)</span>
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Read 2 more answers
2.714 mole of helium has a mass of ______ grams.<br><br> (Enter just the number for this one.)
kifflom [539]

Hello!

To find the mass of helium, we need to multiply the total moles by the mass of helium. We are given 2.714 moles of helium, and the mass of helium is about 4.00 grams. Now, we multiply the two values together to get the grams.

2.714 moles x 4.00 grams = 10.856 grams

According to the number of significant figures, 2.714 moles of helium has a mass of 10.9 grams (exact value: 10.856 grams).

6 0
3 years ago
18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
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