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jonny [76]
2 years ago
8

A chemical reaction is done in the setup shown , resulting in a change of mass. What will happen if the same reaction is done in

a sealed container that is placed on the electronic balance?
Chemistry
2 answers:
FrozenT [24]2 years ago
7 0

Correct Answer: B. The mass will stay the same because the gases produced in the reaction are kept inside the container.

Explanation: The law of conservation of matter states that, in any physical or chemical process, matter is neither created nor destroyed. Because the amount of matter does not change, the mass must remain the same.

The reason the mass decreases in the original reaction is that one of the products of the reaction is a gas, which floats away into the surrounding air. Because the mass of the gas is removed, the mass on the scale goes down. In a closed container, the gas produced in the reaction will not escape, so the total mass inside the container does not change.

irina1246 [14]2 years ago
5 0
<h2>Answer:</h2>

The mass of the system will remain the same if there is no conversion of mass to energy in the reaction.

<h3>Explanation:</h3>
  • If the system is closed, then according to the law of mass conservation the mass of the reaction system will remain the same.
  • <u><em>Law of conservation of the mass: In simple words, it is described as the mass of a closed system can never be changed, it may transfer from one form to another or change into energy.</em></u>
  • But if the reaction involves energy transfer like heat or light production, in this case, the mass can be changed.
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If a 0.10 M solution of a colored substance has a maximum absorbance at 500 nm and an absorbance of 0.26 M at this wavelength, w
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Absorbance is related to the concentration of a substance using the Beer-Lambert's Law. According to this law, absorbance is linearly related to concentration. However, this is only true up to a certain concentration depending on the substance. For this case, we assume that the said law is applicable.

A = kC
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1 year ago
Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b
frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

(a) Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}

(b) 2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}

(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

(d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Answer: (d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

Aluminum's half-equation is oxidation:

Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}

3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}

Adding them:

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

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What is the molarity of 4.35 moles kmno4 dissolved in 750 ml of solution?
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M= #moles / L
 
4.35/.75 = 5.6
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