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miss Akunina [59]
3 years ago
14

the chemical formula for copper iodide is Cul. How many atoms of copper(Cu) is it made from? How many atoms of iodine(I)?

Chemistry
1 answer:
Alexandra [31]3 years ago
7 0
<span>The molecule contains one atom of copper and one atom of iodine. They are connected by an ionic bond because the copper takes a positive charge and the iodine has a negative charge before they are bonded. These opposing charges are negated when the two elements come together.</span>
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Please help with chemistry homework (I’ll mark brainliest!)
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Answer:

You answer is correct

Electrons have a negative charge. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.

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A student was given a stock solution with a concentration of 3.61 g/mL and asked to perform
vovangra [49]

Answer:

The concentration of the dilute sample will be 0.361 g/ml

Explanation:

If a solution is diluted into 1:10 ratio then the amount of solute of that solution will be decreased by 10 times.

      The initial concentration of the stock solution was 3.61g/ml but when the solution is diluted in 1:10 ratio the solute concentration is also decreased by 10 times.SO at present the solute concentration becomes 3.61/10=0.361 g/ml.

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

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