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olya-2409 [2.1K]
3 years ago
12

Under what pressure did 10L of gas change to if the original pressure was 4 atm on 3L?​

Chemistry
1 answer:
cluponka [151]3 years ago
3 0

Answer:

11111

Explanation:

1110001001

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What are the coefficients when the chemical equation below is balanced?
timofeeve [1]

Answer:

C is correct

Explanation:

4 0
3 years ago
list 10 chemical reactions that have benefited your life today. Include the reasons you think each was indeed a chemical reactio
Dennis_Churaev [7]

1) The overall balanced photosynthesis reaction:  

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.

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C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.

Organic carbon (glucose) is transformed into inorganic carbon (carbon dioxide) that goes into atmosphere.

3) Gasoline combustion:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O.

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Gasoline is being burn.

4) Balanced chemical reaction of forming rust: 4Fe + 3O₂ → 2Fe₂O₃.

Forming rust usually last for few months and iron changing in another substance with different properties.

5) An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.  

In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).  

Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.

Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

Copper is forming from the solution, that is chemical change.

6) Balanced chemical reaction: CaO + H₂O → Ca(OH)₂.

Making limewater  (diluted solution of calcium hydroxide).

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4 0
3 years ago
What is an ionic bond? A. the attraction positive ions have for free electrons B. the force that holds the valence electrons to
Olin [163]
Ionic bond is a type of chemical bond that refers to the bonding of <span>oppositely charged ions (anions and cations) because of attraction and the </span>transfer of valence electron(s) between atoms. Cation is the metal that loses electrons and become a positively charged cation, and anions are the nonmetal that accepts those electrons to become a negatively charged anion.
According this explanation, an ionic bond is:
B. the force that holds the valence electrons to the atom


6 0
3 years ago
What is one beneift of using scientific notation instead of regular notation
kotykmax [81]

Answer:

its shorter than a regular one

Explanation:

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8 0
2 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
2 years ago
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