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creativ13 [48]
2 years ago
12

Why is CO2 a pure substance?

Chemistry
1 answer:
allochka39001 [22]2 years ago
7 0
Two or more pure substances can be mixed together to form a mixture. An example of a mixture of two compounds is alcohol and water. An example of a mixture of elements and compounds is air. The major components of air are hydrogen, oxygen, nitrogen and carbon dioxide.
You might be interested in
How can one tell how much of each reactant will be used in a reaction?
Ostrovityanka [42]
To Tell how much of each reactant will be used in a reaction, we need to find which reactant is the Limiting Reagent.
All the reactants will be consumed in equal amount as that of L.R.
5 0
3 years ago
CH3COOH  CH3COO– + H+
Oxana [17]

(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

6 0
2 years ago
Can somewhere show me the dimensional analysis of this?
natita [175]

245 mm Hg = 32.6634 kPa

<u>Explanation:</u>

When the pressure inside the can is measured in mm Hg and it is needed to convert mm mercury (Hg) to kilo pascal, we have to multiply the pressure in mm Hg with  0.13332, so that the pressure is converted in kilo pascals.

1 mm Hg × 0.13332 = 1 kPa

245 mm Hg × 0.13332 = 32.6634 kPa

So the pressure in mm mercury is converted into kilo pascals.

6 0
2 years ago
Identify the false statement from the following.
LUCKY_DIMON [66]

The false statement from the above is that: Temporary charge imbalances in the molecules lead to London dispersion forces.

<h3>What are the factors that affect London dispersion forces?</h3>

Generally, the factors which affects the London dispersion forces a dispersion force are as follows:

  • Shape of the molecules
  • Distance between molecules
  • Polarizability of the molecules

However, London dispersion forces simply refers to a sort of temporary attractive force formed when electrons in two adjacent atoms occupy positions that make the atoms form dipoles.

So therefore, temporary charge imbalances in the molecules lead to London dispersion forces is a false statement

Learn more about London dispersion forces:

brainly.com/question/1454795

7 0
2 years ago
Which term represents the movement of molecules from an area of lower concentration to one of higher concentration with carrier
poizon [28]

Answer:

Active transport

Explanation:

Active transport is defined as the movement of ions or molecules from a region of lower concentration into a region of higher concentration by the use of energy. Two examples of active transport include the uptake of glucose in human intestine and the absorption of mineral ions into plant roots.

Active transport requires energy because it involves the movement of molecules against an existing concentration gradient.

4 0
2 years ago
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