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Andreyy89
2 years ago
14

Explain the relationship between the volume of a gas and it’s pressure.

Chemistry
1 answer:
Olin [163]2 years ago
8 0

Answer:

Boyle's law is a gas law, stating that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If volume increases, then pressure decreases and vice versa, when temperature is held constant

Explanation:

You might be interested in
Which solution whose POH is 12
lesya692 [45]

Answer:

d

Explanation:

H3O =1×10_12=12 ooh is 12

3 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
When Sulphur dioxide reacts with water the product formed is
rosijanka [135]

Answer:

2 .Sulphurous acid

Explanation:

Sulphur dioxide can dissolve in water to form Sulfurous acid(H2SO3). sulphurous acid is weackly dibasic acid. sulphur dioxide is a major component of acid rain since it mixes with vapour in the atmosphere reacting to produce H2So4 .

7 0
3 years ago
Read 2 more answers
A regular 3-d spatial arrangement of points that correspond to atom positions within a crystal is called_______?
Soloha48 [4]

Answer:

Explanation:

Lattice

8 0
1 year ago
Which of the following possess the greatest concentration of hydroxide ions?
jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

11=-\log [OH^-]

[OH^-]=1.0\times 10^{-11}M

Thus, the OH^- concentration is, 1.0\times 10^{-11}M

<u>(b) a solution of 0.10 M NH_3</u>

As we know that 1 mole of NH_3 is a weak base. So, in a solution it will not dissociates completely.

So, the OH^- concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the OH^- concentration.

pOH=-\log [OH^-]

12=-\log [OH^-]

[OH^-]=1.0\times 10^{-12}M

Thus, the OH^- concentration is, 1.0\times 10^{-12}M

<u>(d) a solution of 0.10 M NaOH</u>

As we know that NaOH is a strong base. So, it dissociates to give Na^+ ion and OH^- ion.

So, 0.10 M of NaOH in a solution dissociates to give 0.10 M of Na^+ ion and 0.10 M of OH^- ion.

Thus, the OH^- concentration is, 0.10 M

<u>(e) a 1\times 10^{-4}M solution of HNO_2</u>

As we know that 1 mole of HNO_2 in a solution dissociates to give 1 mole of H^+ ion and 1 mole of NO_2^- ion.

So, 1\times 10^{-4}M of HNO_2 in a solution dissociates to give 1\times 10^{-4}M of H^+ ion and 1\times 10^{-4}M of NO_2^- ion.

The concentration of H^+ ion is 1\times 10^{-4}M

First we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (1.0\times 10^{-4})

pH=4

Now we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

10=-\log [OH^-]

[OH^-]=1.0\times 10^{-10}M

Thus, the OH^- concentration is, 1.0\times 10^{-10}M

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0
3 years ago
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