Answer:
c
Explanation:
the cells grow more than the tissue
Answer:
0.521 moles still present in the container.
Explanation:
It is possible to answer this question by using the general gas law, that is:
PV = nRT
<em>Where P represents pressure of the gas, v its volume, n moles, R gas constant law and T absolute temperature (21.7°C + 273.15 = 294.85K)</em>
Replacing with values of the initial conditions of the container, its volume is:
V = nRT / P
V = 2.00mol*0.082atmL/molK*294.85K / 3.75atm
V = 12.9L
When some gas is released, absolute temperature is 28.1°C + 273.15 = 301.25K, the pressure is 0.998atm and <em>the volume of the container still constant. </em>Again, using general gas law:
PV / RT = n
0.998atm*12.9L / 0.082atmL/molK*301.25K = n
0.521 moles = n
<h3>0.521 moles still present in the container.</h3>
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Answer:
No, Stephanie is incorrect. Formation of petroleum cannot take place under the presence of oxygen.
Explanation:
Since, the petroleum is fossil product. Fossil fuel are formed under high pressure and temperature with absence of oxygen for longer period. so the way she is performing is completely incorrect. With the presence of oxygen in no way petroleum will be formed. The temperature and pressure should be in different combination for the formation of the petroleum. Along with the layers of sediments to maintain the pressure is required.
A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll
The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH
Excess reactant is methanol, 10 mol are used up therefore 24 - 10 mol = 14 mol are remaining at the end of the reaction