Answer:
There are three possible chemical equations for the combustion of sulfur:
- 2S (s) + O₂ (g) → 2SO (g)
- 2S (s) + 3O₂ (g) → 2SO₃ (g)
Explanation:
<em>Combustion</em> is a reaction with oxygen. The products of the reaction are oxides, and energy is released in the form of heat and light.
<em>Sulfur</em> iis a nonmetal, so the oxide formed is a nonmetal oxide.
The most common oxidation numbers of sulfur are -2, + 2, + 4, and + 6.
The combination of sulfur with oxygen may be only with the positive oxidation numbers (+2, + 4, and +6).
Then you have three different equations for sulfur combustion:
<u>1) Oxidation number +2:</u>
Which when balanced is: 2S(g) + O₂(g) → 2SO(g)
<u>2) Oxitation number +4:</u>
That equation is already balanced.
<u>3) Oxidation number +6:</u>
Which when balanced is: 2S(s) + 3O₂(g) → 2SO₃(g)
Answer:
$13,500
Explanation:
The differential analysis of the proposal to replace the commercial oven is shown below:-
The Total differential decrease in cost = Annual maintenance cost reduction × Number of years applicable
= $23,000 × 5
= $115,000
Inflow cash = The Total differential decrease in cost + Proceeds from sale of equipment
= 115,000 + $8,500
= $123,500
The Net differential decrease in cost from replacing equipment = Inflow cash - Cost of new equipment
= $123,500 - $110,000
= $13,500
Respiration aids in replenishing O2 and decrease the CO2 level in the cell. O2 is an important receiver in the oxidative respiration chain in the mitochondria, which is responsible for producing ATP for the cell. ATP can be considered as the “energy currency” of the cell. Many biochemical reactions in the cell requires ATP, including the operations of ion channels, buffer systems, transportation via vesicles, etc. Therefore if there is no O2, the cell cannot produce ATP and will die.
Explanation:
Take shelter in a hard wall building
Close doors and windows cut off ventilation
Answer:
Oxygen is the limiting reactant.
Explanation:
Hello,
In this case, given the reaction:
![C_2H_6O(g) + 3O_2(g)\rightarrow 2CO_2(g) + 3H_2O(g)](https://tex.z-dn.net/?f=C_2H_6O%28g%29%20%2B%203O_2%28g%29%5Crightarrow%202CO_2%28g%29%20%2B%203H_2O%28g%29)
Hence, given the masses of both ethanol and oxygen, we are able to compute the available moles ethanol by:
![n_{C_2H_6O}^{available}=0.461g*\frac{1mol}{46g}=0.01mol C_2H_6O](https://tex.z-dn.net/?f=n_%7BC_2H_6O%7D%5E%7Bavailable%7D%3D0.461g%2A%5Cfrac%7B1mol%7D%7B46g%7D%3D0.01mol%20C_2H_6O)
Next, we compute the moles of ethanol that react with the 0.640 grams of oxygen considering their 1:3 molar ratio in the chemical reaction:
![n_{C_2H_6O}^{consumed\ by\ O_2}=0.64gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6O}{3molO_2} =0.0067molC_2H_6O](https://tex.z-dn.net/?f=n_%7BC_2H_6O%7D%5E%7Bconsumed%5C%20by%5C%20O_2%7D%3D0.64gO_2%2A%5Cfrac%7B1molO_2%7D%7B32gO_2%7D%2A%5Cfrac%7B1molC_2H_6O%7D%7B3molO_2%7D%20%3D0.0067molC_2H_6O)
In such a way, since there are 0.01 available moles of ethanol but just 0.0067 moles are reacting, we evidence ethanol is in excess, therefore the oxygen is the limiting reactant.
Best regards.