<span>Not to be confused with tetration.
This article is about volumetric titration. For other uses, see Titration (disambiguation).
Acid–base titration is a quantitative analysis of concentration of an unknown acid or base solution.
Titration, also known as titrimetry,[1] is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Since volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant or titrator[2] is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrand[3] to determine concentration. The volume of titrant reacted is called titration volume</span>
Answer:
The OH group
Explanation:
Benzhydrol contains OH hydroxyl group in its molecule while fluorene does not. At first glance, one would think that OH, which contributes to hydrogen bonding would causes melting point of benzhydrol to be higher than fluorene. <em>However, </em>the structure of benzhydrol, which is 2 benzene rings connected to center hydroxyl carbon (PhCOHPh), allows for each benzene rings in benzhydrol to rotate until both rings are perpendicular to minimize repulsive force. This prevents the molecule from stacking on each other due to its non flat shape, and thus, lowering its melting point in contrast to flat fluorene molecule.
11.48-gram of 
are needed to produce 6.75 Liters of 
gas measured at 1.3 atm pressure and 298 K
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
First, calculate the moles of the gas using the gas law,
PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.
Given data:
P= 1.3 atm
V= 6.75 Liters
n=?
R= 
T=298 K
Putting value in the given equation:
Moles = 0.3588 moles
Now,
Mass= 11.48 gram
Hence, 11.48-gram of are needed to produce 6.75 Liters of gas measured at 1.3 atm pressure and 298 K
Learn more about the ideal gas here:
brainly.com/question/27691721
#SPJ1
