Answer:
Choice A: approximately 
.
Explanation:
Note that the unit of concentration, 
, typically refers to moles per liter (that is: 
.)
On the other hand, the volume of the two solutions in this question are apparently given in 
, which is the same as 
(that is: 
.) Convert the unit of volume to liters:
.
.
Calculate the number of moles of 
formula units in that 
of the 
solution:
.
Note that (sulfuric acid) is a diprotic acid. When one mole of completely dissolves in water, two moles of 
ions will be released.
On the other hand, 
(sodium hydroxide) is a monoprotic base. When one mole of formula units completely dissolve in water, only one mole of 
ions will be released.
ions and ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid dissolves in water completely, it will take two moles of to neutralize that two moles of produced. On the other hand, two moles formula units of the monoprotic base will be required to produce that two moles of . Therefore, and formula units would neutralize each other at a two-to-one ratio.
.
.
Previous calculations show that 
of was produced. Calculate the number of moles of formula units required to neutralize that
.
Calculate the concentration of a 
solution that contains exactly 
of formula units:
.
Capillary action is defined as the ability of a liquid to go up a narrow space without the help or opposition of external forces. One of the most important factors affecting capillary action is the intermolecular forces within a substance. The higher the IMF, the greater the capillary action. The H-bonding in water gives it greater IMF than acetone, so water has greater capillary action.
Answer:
Absolutely True :) cause we use it all the time
Hi friend
--------------
Your answer
-------------------
Water = H2O
Number of molecules in one mole of water = 6.022 × 10²³ [Avogadro's constant]
Given number of molecules = 2.52 × 10²³
So,
------
Number of moles =
HOPE IT HELPS
Answer:
650 mmol.
Explanation:
The equation for the fermentation of one mole of glucose is:
C₆H₁₂O₆ + 2 NAD⁺ + 2 ADP + 2 P i + 2 NADH → 2 EtOH + 2 ATP + 2 NADH + 2 NAD⁺
Since NAD⁺/NADH is used and regenerated, we can eliminate it from the equation:
C₆H₁₂O₆ + 2 ADP + 2 P i → 2 EtOH + 2 ATP
With the equation, we calculate the maximum amount of ethanol that could be obtained theoretically:
1000 mmol C₆H₁₂O₆ ------------ 2000 mmol EtOH
325 mmol C₆H₁₂O₆ ------------- x= 650 mmol EtOH
Therefore, the maximum amount of ethanol that could be produced is 650 mmol.
