All categories

3 years ago

What is a substance and what are the two types of substances?

Chemistry

2 answers:

vladimir2022 [97]3 years ago

8 0

Answer:

The two types of substances are elements and compounds.

Explanation:

A pure substance is one in which the composition is definitive and constant throughout, and that has distinct chemical properties. These substances are comprised of only one entity, not a mixture of two or more.

Elements are pure substances, assuming they are free of imperfections, because they are made of only one type of atom.

Compounds are also pure substances, like water, table salt, table sugar, etc. because they are comprised of one type of compound (H2O, NaCL, C6H1206,etc)

masha68 [24]3 years ago

6 0

A pure substance has a constant composition and cannot be separated into simpler substances by physical means. There are two types of pure substances: elements and compounds. Elements: are pure substances made up of only l type of atom. Atoms of the same element are identical in properties.

You might be interested in

15) The average human is ~60% water, which translates to ~44.3 kg of water. If you eat a Reece's peanut butter cup (105 Calories

EastWind [94]

Answer: D) 2.37°C

Explanation:

$4.39x10^5J(\frac{g°C}{4.184J} )(\frac{kg}{1000g})(\frac{1}{44.3kg} )$ $=2.37°C$

3 0

2 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

How many molecules are in 4.62 moles of nitric acid (HNO3)?

Lena [83]

Answer:

2.8x10^24

Explanation:

To convert moles to molecules, multiply the number of moles by Avagadro's number (6.02x10^23. Round if required.

4.62mol × 6.02x10^23 = 2.8x10^24

6 0

3 years ago

Need help ASAP!<br><br> How many moles of sodium nitrate are in 0.25 L of 1.2 M NaNO3 solution?

Setler79 [48]

Answer:

$\boxed {\boxed {\sf 0.3 \ mol \ NaNO_3}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

molarity= 1.2 mol NaNO₃/L
liters of solution=0.25 L
moles of solute =x

$1.2 \ mol \ NaNO_3/L= \frac{x}{0.25 \ L}$

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

$0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L$