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AVprozaik [17]
3 years ago
12

How many moles are in 2.5L of 1.75 M Na2CO3

Chemistry
2 answers:
ra1l [238]3 years ago
7 0

Answer:

4.4 mol.

Explanation:

Have a great day

Natasha_Volkova [10]3 years ago
4 0

Answer:

4.4 mol.

Explanation:

Hello!

In this case, since the formula for calculating the molarity is:

M=n/V

Whereas n stands for moles and V for the volume in liters; we can solve for n as shown below when we are given the volume and the molarity:

n=V*M

Thus, we plug in the given data to obtain:

n=2.5L*1.75mol/L=4.4mol

Best regards!

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3 years ago
What is the pH of a solution of RbOH with a concentration of 0.86 M? Answer to 2 decimal places
lubasha [3.4K]

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

Explanation:so you can't use

pH

=

−

log

(

0.150

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because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

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pOH

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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

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

−−−−−−−−−−−−−−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

−−−−−−−−−−−−−−

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<span>

</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq)       pKₐ₂


HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq)          pKₐ₃ 


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