All categories

2 years ago

Examples of pure and impure substances

Chemistry

2 answers:

erastova [34]2 years ago

8 0

Pure- table salt
Impure- vegetable oil

Sladkaya [172]2 years ago

8 0

Examples of pure substances are water, diamond, table salt

Examples of impure substances are muddy water, air, sugar in water

Explanation:

A pure substance is one that is made up of only one element or one chemical compound. It has no impurities. They can be elements or compounds.
It has definite properties and composition.
Examples are water, diamond, table salt, etc.
An impure substance is one that is made up of two or more elements or compounds that are not chemically bonded. It is a mixture.
It has no definite composition or properties.
Examples are muddy water, sugar in water, air, etc.

You might be interested in

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the calculation of the moles of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to calculate the grams of C and O and then do the substraction form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can convert the grams of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the smallest one:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

6 0

3 years ago

How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5

NeTakaya

Answer:

Explanation:

How many moles of gold are equivalent to 1.204 × 1024 atoms?

0.2

0.5

C) 2 Is the correct answer, I took the test and it was correct.

5 0

2 years ago

Read 2 more answers

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices

BaLLatris [955]

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

3 0

3 years ago

The substances that are produced after the chemical reaction are called the ___

sertanlavr [38]

Answer:

Products

Explanation:

During a chemical reaction, there are the reactants (left side), and the products (right side).

6 0

2 years ago

Read 2 more answers

Copper was the first metal to be produced from its ore because it is the easiest to smelt, that is, to refine by heating in the

Ganezh [65]

Answer:

57.48%

Explanation:

Calculate the mass of 1 mole of malachite:

MM Cu = 63.55

MM O = 16.00

MM H = 1.01

MM C = 12.01

$(Cu_{2}(OH)_{2}CO_{3})$

A mole of malachite has:

2 moles of Cu

5 moles of O

2 moles of H

1 mole of C

MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)

MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01

MW Malachite = 221.13

Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1

Now divide the mass of Cu by the mass of Malachite

$%Cu = \frac{127.1}{221.13} =0.5748=57.48%$

7 0

3 years ago