Answer:
used to determine the concentration of the acidic solution by titrating it
Answer:
516.77 grams of Argon gas is present
Explanation:
Using the gas formula
PV = nRT
number of moles (n) = mass / molar weight or mass
P = pressure = 3.4 atm
V = volume = 72 L
R = gas constant = 0.082 L atm mol^-1 K^-1
T = temperature = 225 K
MM = molar mass of Ar = 38.984 g/mol
PV = mRT/ MM
m = PV MM / RT
m = 3.4 * 72 * 38.948 / 0.082 * 225
m = 9534.4704 / 18.45
m = 516.77 grams
the mass of Ar gas you have is 516.77 grams.
CH2OHCH2OH is a
general example of a polyhydroxyl alcohol. A polyhydroxyl alchol is one in which
there are two hydroxyl groups present in the substance. The –OH group attached
to both the carbon atoms.
S + O2 → SO2
<span>z / (32.0655 g S/mol) x (1 mol SO2 / 1 mol S) x (64.0638 g SO2/mol) = (1.9979 z) g SO2 </span>
<span>C + O2 → CO2 </span>
<span>(9.0-z) / (12.01078 g C/mol) x (1 mol CO2 / 1 mol C) x (44.00964 g CO2/mol) = (32.9776 - 3.66418 z) g CO2 </span>
<span>Add the two masses of SO2 and CO2 and set them equal to the amount given in the problem: </span>
<span>(1.9979 z) + (32.9776 - 3.66418 z) = 27.9 </span>
<span>Solve for z algebraically: </span>
<span>z = 3.0 g S</span>
You use a nucleus with inflection on the arm processor.
