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3 years ago

9.56 x 10^5 cg is how many grams?

Chemistry

1 answer:

Liula [17]3 years ago

4 0

Answer:

9.56×10¯⁷ g.

Explanation:

Mass in cg = 9.56×10¯⁵ cg

Mass in g =.?

Thus, we can convert 9.56×10¯⁵ cg to grams (g) as illustrated below:

Recall:

1 cg = 0.01 g

Therefore,

9.56×10¯⁵ cg = 9.56×10¯⁵ cg × 0.01 g / 1 cg

9.56×10¯⁵ cg = 9.56×10¯⁷ g

Therefore, 9.56×10¯⁵ cg is equivalent to 9.56×10¯⁷ g.

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3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

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Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

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We are given:

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Here, tin will undergo reduction reaction and will get reduced.

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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:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

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Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

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