Kilauea volcano in Hawaii emits 200-300 tons of sulfur dioxide into the atmosphere each day. This is an example of
- the impact of natural processes on the earth's environment.
- air pollution from a natural source.
- the magnitude of the chemistry associated with the environment.
Kilauea volcano in Hawaii emits noxious compounds of sulfur dioxide and other harmful pollutants as a result of a reaction with atmospheric water vapors and oxygen.
This reaction results in acid rain and volcanic smog which pollutes the air.
Over the years, the volcano has become a potential threat to health as harmful oxides are accelerating respiratory problems and acid rain destroys crops, and also harms water supplies.
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Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
817.567 mm hg the answer for number 2
Because they can't get trapped in.
Answer:
(40 g O) / (15.99943 g O/mol) x (1 mol CaCO3 / 3 mol O) x (100.0875 g CaCO3/mol) =
83 g CaCO3
So answer D), although three significant digits are not justified.
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