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QveST [7]
2 years ago
11

Determine the mass of the following samples (remember mass => grams)

Chemistry
1 answer:
kakasveta [241]2 years ago
5 0

Mass is the amount of content available in the matter. The mass of the moles of samples given are

2 moles Co = 117.86 g Co

5 moles H2 =  10.10 g  H₂

2.1 moles H₂O = 37.84 g  H₂O

<h3>What is a mole?</h3>

A  mole can be atoms or molecules or ions or any substance.

The mass of one mole of Co is 58.93 g.

Then, 2 moles of Co will be 2 x 58.93 g = 117.86 g Co

The mass of one mole of H₂ is 2.02 g.

Then, 5 moles of  H₂ will be 5 x 2.02 g =  10.10 g  H₂

The mass of one mole of H₂O is 18.02 g .

Then, 2.1 moles of  H₂O will be 2.1 x 18.02 g =  37.84 g  H₂O

Thus, the mass of the following samples in grams

2 moles Co = 117.86 g Co

5 moles  H₂=  10.10 g  H₂

2.1 moles H₂O = 37.84 g  H₂O

Learn more about mole.

brainly.com/question/26416088

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Answer:

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if you need an explanation to each lmk

8 0
3 years ago
An analysis of an ionic compound is found to contain 92 grams of sodium (Na) and 32 grams of oxygen (O). Determine the empirical
Darina [25.2K]
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6 0
3 years ago
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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
Rusting of iron is a very common chemical reaction. It results in one form from Fe reacting with oxygen gas to produce iron (III
Vlada [557]

<u>Answer:</u> The given amount of iron reacts with 9.0 moles of O_2 and produce 6.0 moles of Fe_2O_3

<u>Explanation:</u>

We are given:

Moles of iron = 12.0 moles

The chemical equation for the rusting of iron follows:

4Fe+3O_2\rightarrow 2Fe_2O_3

  • <u>For oxygen gas:</u>

By Stoichiometry of the reaction:

4 moles of iron reacts with 3 moles of oxygen gas

So, 12.0 moles of iron will react with = \frac{3}{4}\times 12.0=9.0mol of oxygen gas

  • <u>For iron (III) oxide:</u>

By Stoichiometry of the reaction:

4 moles of iron produces 2 moles of iron (III) oxide

So, 12.0 moles of iron will produce = \frac{2}{4}\times 12.0=6.0mol of iron (III) oxide

Hence, the given amount of iron reacts with 9.0 moles of O_2 and produce 6.0 moles of Fe_2O_3

5 0
3 years ago
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