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3 years ago

Explain why aluminum does not react with potassium nitrate (KNO3) although it reacts with copper nitrate

Chemistry

1 answer:

Ratling [72]3 years ago

7 0

Answer:

Potassium is more reactive than aluminium, so no reaction takes place. But aluminium is more reactive than copper, so it replaces the copper in copper nitrate

<h3>Explanation:</h3>

More reactive metal compound + less reactive metal

-> no reaction

However

Less reactive metal compound + more reactive metal

-> more reactive metal compound + less reactive metal

This is called substitution reaction where the more reactive metal replaces the less reactive metal in the compound.

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Summarize the process a scientist goes through to come up with a<br> satisfactory solution.

maria [59]

Guess and check, test, trial and error, completion.

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3 years ago

Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C

timurjin [86]

Answer:

The specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-$

Reduction half reaction: $2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-$

Thus, the specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

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4 years ago

A gas has an initial volume of 455 mL at 105ºC and a final volume of 235 mL. What is its final temperature in Celsius degrees?

Oksana_A [137]

Hello!

To solve this problem we're going to use the Charles' Law. This Law describes the relationship between Volume and Temperature in an ideal gas. Applying this law we have the following equation:

$\frac{V1}{T1} = \frac{V2}{T2} \\ \\ T2= \frac{V2*T1}{V1}= \frac{235 mL * 105 ^{\circ}C }{455 mL}=54,23 ^{\circ}C$

So, the final temperature is 54,23 °C

Have a nice day!

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