Figure one illustrates the use of six steel pipes to transport water, dry air, or a
mixture of water and air. The internal surfaces and three of these pipes are coated fully or partially with
another material. An engineer performs an experiment to investigate differences in rusting between the
pipes. The following method is applied to all of the pipes: 1) measure the dry mass of a
pipe using a balance, 2) pass air and/or water through the pipe as illustrated for
one month, 3) measure the dry mass of the pipe after one month. Table one shows the measurements recorded by the engineer.
Recall that pipe A had only water. Pipe B had only air. Pipe C had a mixture of water and air. Pipe D had a mixture of water and air and a full coating of zinc. Pipe E had a mixture of water and air also but only half a coating of zinc. And pipe F had a mixture and water and air with half a coating of paint. What’s flowing through these pipes and how they are coated will be important
What is the resolution of the balance used by the engineer? Tick one box: one times 10 to the minus three kilograms, one times 10 to the minus
two kilograms, one times 10 to the minus one kilograms, or one kilogram.
The resolution of a balance is the smallest change in mass than it can measure. Here we can see that all the values in the table are accurate to two decimal
places. The smallest measurement that the balance can detect is therefore 0.10 kilograms,
which can be rewritten in standard form as one times 10 to the minus two
kilograms. This is therefore the correct answer. If the other answers were correct, we would expect all the balance measurements in
the table to be accurate to 0.001 kilograms, 0.1 kilograms, or one kilograms. We can see from the table that this is not the case, therefore we are sure about our
Pipes C and D both increase in mass during the experiment. Calculate the difference between the percentage increases in mass of the two
pipes. Give your answer to three significant figures.
The first step is to calculate the change in mass for pipes C and F as a result of
the experiment. By subtracting the initial mass of each pipe from its final mass, we get the mass
change. In the case of pipe C, it gained 0.56 kilograms. And in the case of pipe F, it gained 0.27 kilograms. So for C, we would write that as the change in mass of C is equal to 10.57 minus
10.01 is equal to 0.56 kilograms. The percentage change in mass is equal to the change in mass divided by the initial
mass. Be careful here not to use the final mass value. You always measure from the start not the end. This is equal to 0.56 divided by 10.01 multiplied by 100 percent, which is equal to
Likewise, for pipe F, the change in mass of F is equal to 12.25 minus 11.98 is equal
to 0.27 kilograms. The percentage change is equal to 0.27 divided by 11.98 times 100 percent, which is
equal to 2.2538 percent. The percentage difference is equal to 5.5944 minus 2.2538, which is equal to 3.341
percent or 3.34 to three significant figures as asked for in the question. Therefore, the difference between the percentage increases in mass of the two pipes C
and F, to three significant figures, is 3.34 percentage points.
Use the results of the engineer’s experiment to answer the following questions. Which factors affect the rusting of the steel pipes? How effective are internal coatings in preventing the rusting of steel pipes? The first thing we’re going to do is analyze exactly which conditions led to an
increase in the mass of the pipes. To do that, we take the initial mass of each pipe and take it away from the final
mass. So for pipe A, there is zero change in the mass. There was no rusting at all. For pipe B, again, there was zero change in the mass, no rusting whatsoever with dry
air. With pipe D, even though there was air and water, there was an internal coating of
zinc and no rusting occurred, whereas with pipe E, with only half coating of zinc,
there was a mild increase in the mass of 0.16 kilograms.
The best place to start here is the case where most rusting occurred. That’s when the most mass change occurred. That’s with pipe C the greatest change of 0.56 kilograms over the month with both
water and air flowing through the pipe. If you recall, that’s the perfect circumstance for rusting. On the other hand, the three pipes that had no change at all, where there was no
rust, were those where there is only water, pipe A, only air, pipe B, or both water
and air but a full internal coating of zinc in pipe D. By comparing pipes D and E, where D has the full zinc coating and E has the half zinc
coating, we can see that because E gains mass and D doesn’t that the half coating is
only partly effective at stopping rust. Finally, by looking at pipes E and F together, we can see that half coating of paint
is not as effective as half a coating of zinc because F, where it had a half coating
of paint, gained more mass than E, which had half a coating of zinc.
So for our answer, we need to turn all these individual points into sentences. So for our first sentence, the presence of air and water together produce the
greatest amount of rusting as shown by pipe C, which gained the most mass. The presence of air or water alone is not sufficient to cause rusting as shown by
pipes A and B, which gained no mass. Take note here because the structure of these answers is very crucial. We start with a statement and then back it up with evidence. The full zinc coating in pipe D was effective in blocking rusting as it gains no mass
despite both air and water passing through it. The partial coatings of paint pipe F and zinc in pipe E were partially effective at
preventing rusting with the air-water mixture as these pipes gained less mass than
pipe C. The zinc was more effective than the paint as pipe E gained less mass than pipe
F. Zinc acts as a sacrificial coating, protecting the steel.
Aluminium is more corrosion-resistant than steel because it undergoes another
chemical reaction to form an unreactive compound. Complete the word equation for this chemical reaction. So we know from the question that we are not looking at a rusting process involving
air and water, but we are looking at a reaction that occurs at the surface of an
aluminium sheet that protects it from reacting further. Aluminium is quite high up the reactivity series, so it reacts really easily with
oxygen in the air. It forms a protective layer of aluminium oxide: Al2O3. And it is this coating that this question is asking about. So aluminium plus oxygen react to form aluminium oxide.