Answer:
Bacteria are vital in keeping nitrogen cycling through the ecosystem, and nitrogen is vital to plant growth. Without bacteria around to break down biological waste, it would build up. And dead organisms wouldn't return their nutrients back to the system
Explanation:
Answer:
Keep it simple. If all the oxygen contained in the 200 grams of potassium chlorate is produced in the decomposition, then all we have to do is find out how many grams of oxygen are there in the 200 grams. This we can do by calculating the ratio of oxygen mass to the whole. Using 39.1 for potassium, 35.45 for chlorine and 3 times 16, or 48 for the oxygen, we get a total of 122.55 grams per mole for potassium chlorate, of which 48 grams are oxygen. This ratio is 48/122.55. This ratio times the original 200 grams of the compound, gives us 78.34 grams of oxygen produced.
Explanation:
2NH₃(g) + CO₂(g) → CO(NH₂)₂(s) + H₂O(l)
is the balanced equation for the synthesis of urea.
43.8 kJ
<h3>
Explanation</h3>
There are two electrodes in a voltaic cell. Which one is the anode?
The lithium atom used to have no oxygen atoms when it was on the reactant side. It gains two oxygen atoms after the reaction. It has gained more oxygen atoms than the manganese atom. Gaining oxygen is oxidation. As a result, lithium is being oxidized.
Oxidation takes place at the anode of a cell. Therefore, the anode of this cell is made of lithium.
Lithium has an atomic mass of 6.94. Each gram of Li would contain 1/6.94 = 0.144 moles of Li atoms. Each Li atom loses one electron in this cell. Therefore, the number of electron transferred, <em>n</em>, equals 0.144 moles for each gram of the anode.
Let 
represents the electrical energy produced.
, where
- <em>n</em> is the <em>number of moles</em> electrons transferred,
- <em>F</em> is the Faraday's constant,
- <em>E</em>
is the cell potential,
<em>n </em>= 0.144 mol, as shown above, and
<em>F </em>= 96.486 kJ / (
).
Therefore,
.
Answer:
At the equivalence point, equal amounts of H+ and OH– ions will combine to form H2O, resulting in a pH of 7.0 (neutral). The pH at the equivalence point for this titration will always be 7.0, note that this is true only for titrations of strong acid with strong base.
Explanation:
