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3 years ago

Write .. hot copper metal reacts with chlorine gas to form solid green copper chloride

Chemistry

1 answer:

maw [93]3 years ago

4 0

Answer:

hope helps.

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Describe how elements are laid out on the periodic table of elements.

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Answer:

The chemical elements are arranged in order of increasing atomic number.

Explanation:

Hydrogen = 1

Helium = 2

Lithium = 3

8 0

3 years ago

Read 2 more answers

What would be the answer

ser-zykov [4K]

Answer:

Explanation:

The balanced equation for the reaction is as follows:

4Al + 3O2 --> 2(Al2O3)

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3 years ago

Use the following half-reactions to construct a voltaic cell:

velikii [3]

<u>Answer:</u> The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

<u>Explanation:</u>

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

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3 years ago

Sulphur burnt in oxygen forms sulphur dioxide. Calculate the heat given out when 1.6g of sulphur is burnt.

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Answer:

Dont know

Explanation:

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3 years ago

The density of the water is 1.00 g/mL.

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Answer:it will sink in water

Explanation:

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3 years ago