(1)Plastics (i.e. synthetic polymers) are used frequently for many applications because they tend to be (a) easy to fabricate items with, by moulding, or other methods, (b) can be strong even though very lightweight, so useful for making items that require good physical strength, as well as things like packaging which need to be able to support the contents correctly, (c) relatively low cost (especially overall, when ease of fabrication is taken into account.
(2)The "use" of plastics is not harmful per se, and plastics themselves are innocuous (one reason that they fail to readily biodegrade is that they do not interact with biological systems). But poor manangement of waste plastic can lead to problems.
(3)Failing to recycle and reuse synthetic polymers can lead to those items not landfilled (which is a separate issue - ultimately we would run out of landfill sites) items ending up in the sea, rivers or in the general environment where they can be hazardous to wildlife, and look unsightly.
But that is down to poor management, and societal attitudes, not down to the fact that the items are made of plastics.
(4)If we failed to manage sewage and simply discharged that into the streets or into rivers that would be a major problem too. But that doesn't mean we should "ban sewage" - just manage it in an effective way.
Generally, the top elements in group 16 will have a charge of -2, although the entire group isn't standardized.
Answer:
The reactants for photosynthesis are light energy, water, carbon dioxide and chlorophyll, while the products are glucose (sugar), oxygen and water.
Explanation:
(a) Formula that shows relation between 
and 
is as follows.
Here, 
= 1
Putting the given values into the above formula as follows.
= 
= 
= 0.01316
(b) As the given reaction equation is as follows.
As there is only one gas so
,
![p[CO_{2}] = K_{p}](https://tex.z-dn.net/?f=p%5BCO_%7B2%7D%5D%20%3D%20K_%7Bp%7D)
= 1.20
Therefore, pressure of 
in the container is 1.20.
(c) Now, expression for for the given reaction equation is as follows.
![K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCaO%5D%5BCO_%7B2%7D%5D%7D%7B%5BCaCO_%7B3%7D%5D%7D)
= 
= \frac{x^{2}}{(a - x)}[/tex]
where, a = initial conc. of 
= 
= 0.023 M
0.0131 = 
x = 0.017
Therefore, calculate the percentage of calcium carbonate remained as follows.
% of remained = 
= 75.46%
Thus, the percentage of calcium carbonate remained is 75.46%.
