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3 years ago

Calculate the volume of a tin block that has a mass of 95.04 grams at stp

Chemistry

2 answers:

stiks02 [169]3 years ago

8 0

Answer : The volume of tin block is, 13.0 mL

Explanation :

Density : It is defined as the mass contained per unit volume.

Formula used for density :

$Density=\frac{Mass}{Volume}$

Given :

Mass of tin = 95.04 grams

Density of tin at STP = 7.287 g/mL

Now put all the given values in the above formula, we get the volume of tin block.

$7.287g/mL=\frac{95.04g}{Volume}$

$Volume=\frac{95.04g}{7.287g/mL}=13.0mL$

Therefore, the volume of tin is 13.0 mL

sergejj [24]3 years ago

3 0

Thank you for posting your question here. I am just assuming to that the density of tin is 7.31 g/cm^3, you can change it it is not the number. Below is the solution. I hope it helps.

95.04g x (1 cm³ / 7.31g) = 13.0 cm³

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In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

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