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r-ruslan [8.4K]
3 years ago
7

What are the steps to naming simple ionic compounds

Chemistry
2 answers:
son4ous [18]3 years ago
8 0
To name an ionic compound, you simply need to find the names of the cation and anion present in the compound and make sure to revise the ends of metal names as needed. First, write out the name of the metal, followed by the name of the non-metal with its new ending.
Olegator [25]3 years ago
3 0

Answer:

To name an ionic compound, you simply need to find the names  of the cation and anion present in the compound and make sure to revise the ends of metal names as needed. First, write out the name of the metal, followed by the name of the non-metal with its new ending.

Explanation:

Write the name of the cation, and then write the name of the anion, changing the ending to "-ide" if it's a single element and leaving it as it is if it's a polyatomic ion.

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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

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

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pH

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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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lubasha [3.4K]
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m = 0.0924 grams

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