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3 years ago

List 4 natural sources of acid rain please help

Chemistry

2 answers:

solong [7]3 years ago

7 0

1. sulfur dioxide
<span>2. </span><span>nitrogen oxides
Sorry that's all i know sorry I hope it still helped</span>

Alexxandr [17]3 years ago

4 0

Acid rain is caused by both natural sources and man-made sources. Natural sources like volcanic eruptions, decaying vegetation, sulfur dioxide gases, and nitrogen oxide gases can mix with water, oxygen and sunlight in the atmosphere that can then cause acid rain. Man made resources are from fossil fuel combustion.

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How many moles of methane are produced when 4.53 mol co2 reacts

ArbitrLikvidat [17]

The reaction formula CH4 + 2O2 → CO2 + 2H2O shows the oxidation of 1 mole of CH4 (Methane) will yield 1 mole of CO2 (Carbon Dioxide). Since 1 mole of CH4 will weigh 12g (for the Carbon) + 4g (1g for each Hydrogen) = 16g, then 32g of CH4 will correspond to 32g / 16g/mole = 2 moles. Therefore the oxidation of 2 moles of CH4 will yield 2 moles of CO2.

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2 years ago

Read 2 more answers

Please let me know if this is correct?

siniylev [52]

Answer: Yes! you're all good. Alkali metals in group 1 are the most metallic :)

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3 years ago

By means of a schematic diagram show how a bacteria cell applied to the region of a cowpea root can end up becoming a nitrate io

GenaCL600 [577]

Answer:

Nitrifying Bacteria are a group of aerobic bacteria important in the nitrogen cycle as converters of soil ammonia to nitrates, compounds usable by plants. An example is nitrosomonas or nitrobacter and species in that family.

The schematic diagram is attached below, which summarises the oxidation of ammonia or free nitrogen in the soil to nitrates for the cowpea plant's utilisation.

4 0

3 years ago

If you have 547.3 grams of Ni2O, how many molecules would be present?

jasenka [17]

Answer: 8.830418848725065

Explanation:

8.830418848725065

5 0

2 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

8 0

3 years ago