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AnnZ [28]
3 years ago
11

Which description is an example of point-source pollution? A. dumping acid from a bucket into a lake B. runoff of insecticide fr

om a field C. pesticides flowing out of people's yards during a rainstorm D. acids that are contained in air settling on a lake surface
Chemistry
2 answers:
Alecsey [184]3 years ago
5 0
A dumping acid from a bucket into a lake

alina1380 [7]3 years ago
5 0

Point source pollution:

Point source pollution is an single source of pollution that is identifiable.

So this source can be different for different pollution that is air, water, land and noise.

Non-point source pollution:

Non-point source pollution is a pollution that results from many sources.

In the given lists of pollution only option A founds to be point source of pollution, that is dumping acid from a bucket into a lake. There is a defined single source of pollution that is acid in the bucket so it is an example of point source of pollution.



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How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci
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Approximately 0.08\; \rm mol, assuming that this gas is an ideal gas.

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Look up the standard room temperature and pressure:25\; \rm ^{\circ}C and P = 101.325 \; \rm kPa.

The question states that the volume of this gas is V = 2\; \rm dm^{3}.

Convert the unit of all three measures to standard units:

\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}.

\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}.

\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}.

Look up the ideal gas constant in the corresponding units: R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}.

Let n denote the number of moles of this gas in that V = 2\; \rm dm^{3}. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

P \cdot V = n \cdot R \cdot T.

Rearrange this equation and solve for n:

\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}.

In other words, there is approximately 2\; \rm mol of this gas in that V = 2\; \rm dm^{3}.

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3 years ago
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