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eimsori [14]
2 years ago
11

A gas occupies 12.3 L at a temperature of 40.0 K. What is the volume when the temperature is increased to 60.0 K?

Chemistry
1 answer:
dimaraw [331]2 years ago
7 0

Answer:

the volume is 18.45 L

Explanation:

The computation of the volume when the temperature is increased to 60.0 K is shown below:

Since the gas occupied 12.3 L at a temperature of 40.0 k

And, the volume when the new temperature is 60.0 k

So, the volume is

= 60.0 k × 12.3 L ÷ 40.0 k

= 18.45 L

hence,  the volume is 18.45 L

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43.8 kJ

<h3>Explanation</h3>

There are two electrodes in a voltaic cell. Which one is the anode?

The lithium atom used to have no oxygen atoms when it was on the reactant side. It gains two oxygen atoms after the reaction. It has gained more oxygen atoms than the manganese atom. Gaining oxygen is oxidation. As a result, lithium is being oxidized.

Oxidation takes place at the anode of a cell. Therefore, the anode of this cell is made of lithium.

Lithium has an atomic mass of 6.94. Each gram of Li would contain 1/6.94 = 0.144 moles of Li atoms. Each Li atom loses one electron in this cell. Therefore, the number of electron transferred, <em>n</em>, equals 0.144 moles for each gram of the anode.

Let w_\text{max} represents the electrical energy produced.

w_\text{max} = n \cdot F \cdot E_\text{cell}, where

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<em>n </em>= 0.144 mol, as shown above, and

<em>F </em>= 96.486 kJ / (\text{V} \cdot \text{mol} \; \text{e}^{-}).

Therefore,

w_\text{max} = n \cdot F \cdot E\\\phantom{w_\text{max}} = 0.144 \times 96.486 \times 3.15 \\\phantom{w_\text{max}} = 43.8 \; \text{kJ}.

3 0
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Based on the equations below, which metal is the least active? Pb(NO3)2(aq) + Ni (s) --&gt; Ni(NO3)2 (aq)+ Pb(s) Pb(NO3)2(aq) +
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Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.

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