I believe that it is liquid.
<span>342.15g/mol I'm not for sure though </span>
Answer:- There are particles in the container present at 23 degree C and 101.3 pKa.
Solution:- The given volume of the container is 4.0 L at 23 degree C and 101.3 kPa pressure.
We would calculate the moles using ideal gas law equation(PV = nRT) and then the moles are converted to particles on multiply the moles by Avogadro number.
Let's convert degree C to kelvin and pKa to atm.
T = 23 + 273 = 296 K
P = 101.3 pKa = 1 atm
R =
V = 4.0 L
n = ?
Let's plug in the values in the equation:
1(4.0) = n(0.0821)(296)
4.0 = n(24.3)
n = 0.165 mol
Let's multiply the moles by Avogadro number to get the number of particles:
= particles
So, from above calculations there would be particles in the container.
Answer:
5.28 moles of NaOH.
Explanation:
Step 1:
Determination of the number of mole in
375.4 g of any Na2S04.
Molar mass of Na2SO4 = (2x23) + 32 + (16x4) = 142g
Mass of Na2SO4 = 375.4g
Number of mole of Na2SO4 =..?
Mole = Mass /Molar Mass
Number of mole of Na2SO4 = 375.4/142 = 2.64 moles
Step 2:
The balanced equation for the reaction.
H2SO4 + 2NaOH —> Na2SO4 + 2H2O
Step 3:
Determination of the number of mole of NaOH needed for the reaction.
From the balanced equation above,
2 moles of NaOH reacted to produce 1 mole of Na2SO4.
Therefore, Xmol of NaOH will react to produce 2.64 moles of Na2SO4 i.e
Xmol of NaOH = 2 x 2.64
Xmol of NaOH = 5.28 moles
Therefore, 5.28 moles of NaOH is needed for the reaction.