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jarptica [38.1K]
2 years ago
15

3. How many moles of solute are present in a 2300 mL solution of 0.68 M MgSO.?​

Chemistry
1 answer:
Nostrana [21]2 years ago
8 0

There are 1.56 moles of solute present in a 2300 mL solution of 0.68M MgSO4.

<h3>How to calculate number of moles?</h3>

The number of moles of a substance can be calculated by using the following formula:

molarity = no. of moles / volume

According to this question, a volume of 2300 mL solution is contained in 0.68 M MgSO4. The number of moles is calculated as follows:

no of moles = 0.68M × 2.3

no. of moles = 1.56

Therefore, there are 1.56 moles of solute present in a 2300 mL solution of 0.68M MgSO4.

Learn more about moles at: brainly.com/question/12127540

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lubasha [3.4K]

Answer:The pH of the solution is given by pH=−log([H3O+])

Explanation:so you can't use

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because that's the concentration of the hydroxide anions,

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Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

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:

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mole ratio.

NaOH

(

a

q

)

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(

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q

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So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

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pOH

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−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

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=

0.824

Now, an aqueous solution at

25

∘

C

has

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=

14

−−−−−−−−−−−−−−so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−

3 0
2 years ago
Methane can be decomposed into two simpler substances: hydrogen and carbon. Therefore, methane 1. is a gas. 2. cannot be an elem
Reika [66]

Answer: 2. cannot be an element.

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CH_4\rightarrow C+2H_2

Thus as it can be decomposed into its constituent elements, it is a compound.

Mixture is a substance which has two or more components which do not combine chemically and do not have any fixed ratio in which they are present.

Thus methane cannot be an element.

3 0
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Leni [432]
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JulsSmile [24]

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

3 0
3 years ago
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