Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
Answer:
It is a salt of two polyatomic ammonium ions and one polyatomic sulfate ion
Explanation:
Ammonium Sulfate is an ionic compound formed by two polyatomic ions, Ammonium NH+4 and Sulfate SO−4 . ... Therefore, it will take two +1 ammonium ions to balance the one -2 sulfate ion. This will make the formula for Ammonium Sulfate (NH4)2SO4
v=fw (Assume for this example w is wavelength). w=v/f. w=100/1000= 0.1 m. The wavelength is 0.1 meters
Answer:
The volume of the gas sample at standard pressure is <u>819.5ml</u>
Explanation:
Solution Given:
let volume be V and temperature be T and pressure be P.



1 torr= 1 mmhg
42.2 torr=42.2 mmhg
so,


Now
firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

=735-42.2=692.8 mmhg
Now
By using combined gas law equation:



Here
are standard pressure and temperature respectively.
we have

Substituting value, we get

