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docker41 [41]
3 years ago
5

Elements, compounds, and mixtures​

Chemistry
1 answer:
nikklg [1K]3 years ago
8 0
Density=mass/volume
1.25=110g/v
v=88mL
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How many liters of water do we need to add to 5.00 moles to get a 0.648 M solution?
Tanya [424]

Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.

Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric

H

3

O

+

.

Moles of nitric acid:

26.0

×

10

−

3

⋅

L

×

8.00

⋅

m

o

l

⋅

L

−

1

=

0.208

⋅

m

o

l

H

N

O

3

(

a

q

)

.

And, moles of hydrochloric acid:

88.0

×

10

−

3

⋅

L

×

5.00

⋅

m

o

l

⋅

L

−

1

=

0.440

⋅

m

o

l

H

C

l

(

a

q

)

.

This molar quantity is diluted to

1.00

L

. Concentration in moles/Litre =

(

0.208

+

0.440

)

⋅

m

o

l

1

L

=

0.648

⋅

m

o

l

⋅

L

−

1

.

Now we know that water undergoes autoprotolysis:

H

2

O

(

l

)

⇌

H

+

+

O

H

−

. This is another equilibrium reaction, and the ion product

[

H

+

]

[

O

H

−

]

=

K

w

. This constant,

K

w

=

10

−

14

at

298

K

.

So

[

H

+

]

=

0.648

⋅

m

o

l

⋅

L

−

1

;

[

O

H

−

]

=

K

w

[

H

+

]

=

10

−

14

0.648

=

?

?

p

H

=

−

log

10

[

H

+

]

=

−

log

10

(

0.648

)

=

?

?

Alternatively, we know further that

p

H

+

p

O

H

=

14

. Once you have

p

H

,

p

O

H

is easy to find. Take the antilogarithm of this to get

[

O

H

−

]

.

Answer link  

4 0
2 years ago
A solution contains some or all of the ions Cu2+,Al3+, K+,Ca2+, Ba2+,Pb2+, and NH4+. The following tests were performed, in orde
xeze [42]

Answer:

See below explanation

Explanation:

When having a mixture of metals in solution, you may perform an analytical study (using selective chemical conditions), that may help you to determine whether a metal (cation) is present or not

Using selective analytes (or conditions), leads to consecutive precipitations, until most of the cations are separated in precipitates

With this technique, you may identify metals in different groups, each group will have its analyte (or condition), which will help to have a different precipitate:

- Group I: Ag⁺, Pb⁺², Hg⁺²;  Analyte: HCL ; Precipitate: AgCl (white) , PbCl₂, HgCl₂

- Group II: As⁺³ , Bi⁺³, Cd⁺², Cu⁺² , Sb⁺³, Sn⁺² ; Analyte: H₂S (g) with HCL ; Precipitate: As₂S₃ , Bi₂S₃ , CdS (yellow) , CuS (black), Sb₂S₃, SnS

- Group III: Co⁺², Fe⁺², Fe⁺³, Mn⁺², Ni⁺², Zn⁺², Al⁺³, Cr⁺³; Analyte: NaOH or NH₃ with (NH₄)₂S (ac) ; Precipitate: CoS (black) , FeS, MnS , NiS (black), ZnS (white) , Al(OH)₃ (white), Cr(OH)₃  

- Group IV: Mg⁺², Ca⁺², Sr⁺², Ba⁺²; Analyte: Na₂CO₃ (ac) or (NH₄)₂HPO₄ (ac); Precipitate: respective carbonate or phosphate MgCO₃/MgHPO₄, CaCO₃/CaHPO₄ , SrCO₃/SrHPO₄, BaCO₃/BaHPO₄

- Group V: Li⁺, K⁺, Na⁺, Rb⁺, Cs⁺, NH₄⁺ ; will remain all in final solution

According to the original statement:

A solution contains one or more of the following: Cu⁺², Al⁺³, K⁺, Ca⁺², Ba⁺², Pb⁺², NH₄⁺

1) Addition on HCl 6M produces no change: we can say the sample does not contain Pb⁺² (group I)

2) Addition of H₂S with 0.2 M HCL produced a black solid: we could say sample contains Cu⁺²(group II)

3) Addition of (NH₄)₂HPO₄ in NH₃ produces no reaction: we could say we don´t have Ca⁺² and /or Ba⁺²  (group IV)

4) The final supernatant, when heated produced a purple flame: in the final solution, we have K⁺ (group V), which produces a purple flame (based on its characteristic emission spectrum when subjected to flame)

This analysis will be inconclusive for NH₄⁺ (according to above describe technique)

6 0
3 years ago
Which of these changes occurs during a nuclear change?
Andrews [41]
The correct answer is D. Elements are created that differ from the reactants.. Because in a eeaction, the product is different than the reactants. Take NaCl for example.
7 0
3 years ago
Why can groundwater only collect in rock that is porous?
Galina-37 [17]

Answer: Porous rock has gaps between rock particles that can fill with water. but i’m not 100% sure

Explanation:

5 0
2 years ago
Read 2 more answers
What should be the temperature of the solvent before adding it to the sample to be recrystallized?
Serhud [2]

Answer:

Near the boiling point of the solvent

Explanation:

The process of recrystallization is hinged on the fact that the amount of solute that can be dissolved by a solvent increases with temperature. The process involves creation of a solution by dissolving a solute in a solvent at or near its boiling point.  At the boiling point of the solvent, the solute has a greater solubility in the solvent; not much volume of  the hot solvent is required to dissolve the solute.

Before the solution is later cooled, you can now filter out insoluble impurities from the hot solvent. The quantity of the original solute drops appreciably because impurities have been removed. At this lower temperature, the solution becomes saturated and the solute can no longer be held in solution hence it forms pure crystals of solute, which can be recovered.

Recrystallization must be carried out using the proper solvent. The solute must be relatively insoluble in the solvent at room temperature but more soluble in the solvent at elevated temperature.

3 0
3 years ago
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