Chapter 5: Problem 5
Why is mercury a more suitable substance to use i a barometer than water?
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 5: Problem 5
Why is mercury a more suitable substance to use i a barometer than water?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeSome commercial drain cleaners contain two components: sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: $$2 \mathrm{NaOH}(a q)+2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow_{2 \mathrm{NaAl}(\mathrm{OH})_{4}(a q)+3 \mathrm{H}_{2}(g)}$$ The heat generated in this reaction helps melt away obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of \(\mathrm{H}_{2}\) formed at STP if \(3.12 \mathrm{~g}\) of \(\mathrm{Al}\) is treated with an excess of \(\mathrm{NaOH}\).
The volume of a sample of pure HCl gas was \(189 \mathrm{~mL}\) at \(25^{\circ} \mathrm{C}\) and \(108 \mathrm{mmHg}\). It was completely dissolved in about \(60 \mathrm{~mL}\) of water and titrated with an \(\mathrm{NaOH}\) solution; \(15.7 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution were required to neutralize the \(\mathrm{HCl}\). Calculate the molarity of the \(\mathrm{NaOH}\) solution.
A gas evolved during the fermentation of glucose (wine making) has a volume of \(0.78 \mathrm{~L}\) when measured at \(20.1^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\) What was the volume of this gas at the fermentation temperature of \(36.5^{\circ} \mathrm{C}\) and 1.00 atm pressure?
If the maximum distance that water may be brought up a well by a suction pump is \(34 \mathrm{ft}(10.3 \mathrm{~m}),\) how is it possible to obtain water and oil from hundreds of feet below the surface of Earth?
At a certain temperature the speeds of six gaseous molecules in a container are \(2.0 \mathrm{~m} / \mathrm{s}, 2.2 \mathrm{~m} / \mathrm{s}, 2.6 \mathrm{~m} / \mathrm{s}\) \(2.7 \mathrm{~m} / \mathrm{s}, 3.3 \mathrm{~m} / \mathrm{s},\) and \(3.5 \mathrm{~m} / \mathrm{s} .\) Calculate the root- mean-square speed and the average speed of the molecules. These two average values are close to each other, but the root-mean-square value is always the larger of the two. Why?
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