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MrRa [10]
3 years ago
11

Please help ASAP I need 1 2 and 3

Chemistry
1 answer:
Radda [10]3 years ago
3 0

Answer:

1 ) C

2) B

3) I'm not sure but I have a feeling its C

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What concentration of the barium ion, ba2 , must be exceeded to precipitate baf2 from a solution that is 1. 00×10−2 m in the flu
Anon25 [30]

The amount of barium ions that must be present in order for the salt to precipitate is 0.245 M.

A solution's solubility product is the result of raising each ion's concentration to the power of its stoichiometric ratio. It is portrayed as

A combination of 1 barium ion and 2 fluoride ions results in the ionic compound known as barium fluoride.

The following equation describes the equilibrium reaction for barium fluoride ionization:

BaF₂ → Ba²⁺ + 2F⁻

Ksp = [Ba²⁺] · [F⁻]²

2.45*10^{-5}= [Ba²⁺] * [1. 00*10^{-2} ]^{2}

[Ba²⁺]=0.245 M

As a result, 0.245 M of barium ions must be present in order for the salt to precipitate.

<h3>Solubility </h3>

Solubility in chemistry refers to a chemical's capacity to dissolve in another substance, the solvent, to produce a solution. Inability of the solute to create such a solution is the opposite quality, or insolubility. A substance's degree of solubility in a given solvent is often determined by the amount of the solute present in a saturated solution, which is a solution in which no additional solute can be dissolved. The solubility equilibrium between the two compounds is considered to have been reached at this time. If there is no such restriction for a given solute and solvent, the two are referred to as being "miscible in any amounts."

What concentration of the barium ion, ba2 , must be exceeded to precipitate baf2 from a solution that is 1. 00×10−2 m in the fluoride ion, f−? ksp for barium fluoride is 2. 45×10−5

Learn more about solubility here:

brainly.com/question/8591226

#SPJ4

4 0
2 years ago
Complete and balance the reaction in acidic solution. equation: ZnS + NO_{3}^{-} -&gt; Zn^{2+} + S + NO ZnS+NO−3⟶Zn2++S+NO Which
kotykmax [81]

Answer:

Balanced reaction: 3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O

S is oxidized and N is reduced.

NO_{3}^{-} is the oxidizing agent and ZnS is the reducing agent.

Explanation:

Reaction: ZnS+NO_{3}^{-}\rightarrow Zn^{2+}+S+NO

\Rightarrow Oxidation: ZnS\rightarrow Zn^{2+}+S

Balance charge: ZnS-2e^{-}\rightarrow Zn^{2+}+S  ...............(1)

\Rightarrow Reduction: NO_{3}^{-}\rightarrow NO

Balance H and O in acidic medium : NO_{3}^{-}+4H^{+}\rightarrow NO+2H_{2}O

Balance charge: NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O ...............(2)

[3\timesEquation-(1)] + [ 2\timesEquation-(2)]:

3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O

Oxidation number of S increases from (-2) to (0) for the conversion of ZnS to S. Therefore S is oxidized.

Oxidation number of N decreases from (+5) to (+2) for the conversion of NO_{3}^{-} to NO. Therefore N is reduced.

NO_{3}^{-} consumes electron from ZnS. Therefore NO_{3}^{-} is the oxidizing agent and ZnS is the reducing agent.

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Elden [556K]

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B or C

Explanation:

I think it's C

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Eyes, nose, mouth would each be an example of a(n)
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Answer:

I think it is Organ

Explanation:

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Explanation:

6 0
3 years ago
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