To be considered acid rain, precipitation would have to

2 answers:

6 0

The two primary sources of acid rain are sulfur dioxide (SO2), and oxides of nitrogen (NOx). Sulfur dioxide is a colourless, prudent gas released as a by-product of combusted fossil fuels containing sulfur. A variety of industrial processes, such as the production of iron and steel, utility factories, and crude oil processing produce this gas. In iron and steel production, the smelting of metal sulfate ore, produces pure metal. This causes the release of sulfur dioxide. Metals such as zinc, nickel, and copper are commonly obtained by this process. Sulfur dioxide can also be emitted into the atmosphere by natural disasters or means. This ten percent of all sulfur dioxide emission comes from volcanoes, sea spray, plankton, and rotting vegetation. Overall, 69.4 percent of sulfur dioxide is produced by industrial combustion. Only 3.7 percent is caused by transportation. Read more at: https://sites.google.com/site/acidrain1project/

irina1246 [14]3 years ago

5 0

Answer: precipitation will have to be deposited with a high level of sulphuric and nitric components

Explanation: precipitation is considered to be acidic when its pH falls below 5.6 (which is 25 times more acidic than pure distilled water). For an acid rain to occur, precipitation will have an acidic component such as sulfuric or nitric acid that fall to the ground from the atmosphere in wet or dry forms.

This can include rain, snow, fog, hail or even dust that is acidic.

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A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.

Gala2k [10]

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The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-

nikdorinn [45]

Answer:

The answer is " $60.74^{\circ}$ ".

Explanation:

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$\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\$

$\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\$