Answer:
2
Explanation:
There are 3 moles O2 in 1 mole CoH12O6 so 2 moles are consumed
Answer:
a) [A⁻]/[HA] = 0.227
b) [A⁻]/[HA] = 0.991
c) [A⁻]/[HA] = 2.667
Explanation:
In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:
pH = pka + Log [A⁻]/[HA]
pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]
4.23 = 4.874 + Log [A⁻]/[HA]
-0.644 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.227 = [A⁻]/[HA]
4.87 = 4.874 + Log [A⁻]/[HA]
-0.004 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.991 = [A⁻]/[HA]
5.30 = 4.874 + Log [A⁻]/[HA]
0.426 = Log [A⁻]/[HA]
= [A⁻]/[HA]
2.667 = [A⁻]/[HA]
Answer:
Yes
Explanation:
Denatured ethanol fuel is a polar solvent, which is soluble in water. A
Polar solvent is a compound with a charge separation in chemical bonds, such as alcohol, most acids, or ammonia. These have affinity with water and will dissolve easily. Denatured fuel ethanol has a flash point of -5 ° F and a vapor density of 1.5, indicating that it is heavier than air.
Consequently, ethanol vapors do not rise, similar to the gasoline vapors they are looking for lower altitudes. The specific gravity of denatured fuel ethanol is 0.79, which indicates that it is lighter than water and has a self-ignition temperature of 709 ° F and a boiling point of 165-175 ° F. Like gasoline, the most denatured fuel, the greatest danger of ethanol as an engine fuel component is its flammability.
It has a wider flammable range than gasoline (LEL is 3% and UEL is 19%).
Answer:
the number of carbon and hydrogen atoms present in the molecule of Ethane that is it contains two carbon atoms and six hydrogen atoms
Explanation:
Answer: 9.3 x 10^ 18 g CO
Explanation:
Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:
(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO
This cancels molecules CO.
Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:
(1 mol CO) x 28 g CO / 1 mol CO
This cancels mol CO, which leaves grams CO.
