Answer:
Amount of sodium sulphate is 5.64g
Amount of water is 1.017g
Explanation:
First give a balanced equation for the reaction
2NaOH + H2SO4 > Na2SO4 + 2H2O
Find the number of moles of each reacting specie by dividing the mass by their molar mass
Use the number of moles to find the amount in the product
For 2NaOH
2.26/80=0.02825moles
0.02825 x molar mass of 2H2O = 1.017g
For H2SO4
4.19/98=0.0428moles
0.0428 x 132 = 5.64g
Answer:
6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.
Explanation:
![Rate = k[AB]^2](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BAB%5D%5E2)
The order of the reaction is 2.
Integrated rate law for second order kinetic is:
Where,
is the initial concentration = 1.50 mol/L
is the final concentration = 1/3 of initial concentration =
= 0.5 mol/L
Rate constant, k = 0.2 L/mol*s
Applying in the above equation as:-
![\frac{1}{1.5}+0.2x=\frac{1}{0.5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1.5%7D%2B0.2x%3D%5Cfrac%7B1%7D%7B0.5%7D)
![t = 6.66\ s](https://tex.z-dn.net/?f=t%20%3D%206.66%5C%20s)
<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>
Answer:
The molar mass of the unknown compound is 202.40 g/mol.
Explanation:
Let the molar mass of compound be M.
The depression in freezing point is given by:
![\Delta T_f=K_f\times m](https://tex.z-dn.net/?f=%5CDelta%20T_f%3DK_f%5Ctimes%20m)
![\Delta T_f=T-T_f](https://tex.z-dn.net/?f=%5CDelta%20T_f%3DT-T_f)
where,
= change in boiling point = 0.81 K
=Boiling point of the solution = 3.77°C
T = freezing point of the pure solvent here benzene =5.48°C
= freezing point constant = 5.12°C/m
m = molality =![\frac{\text{Mass of solute}}{\text{Molar mass of solute}\text{Mass of solvent in kg}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctext%7BMass%20of%20solvent%20in%20kg%7D%7D)
![\Delta T_f=5.48^oC-3.77^oC=1.71^oC](https://tex.z-dn.net/?f=%5CDelta%20T_f%3D5.48%5EoC-3.77%5EoC%3D1.71%5EoC)
![1.71^oC=5.12^oC/m\times \frac{33.8 g}{M\times 0.500 kg}](https://tex.z-dn.net/?f=1.71%5EoC%3D5.12%5EoC%2Fm%5Ctimes%20%5Cfrac%7B33.8%20g%7D%7BM%5Ctimes%200.500%20kg%7D)
M = 202.40 g/mol
The molar mass of the unknown compound is 202.40 g/mol.
<span>1.67 x 10^-3 moles
Calculate the molar mass of C9H8O4
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
C9H8O4 = 9 * 12.0107 + 8 * 1.00794 + 4 * 15.999 = 180.1558
Now divide the mass you have by the molar mass
0.300 g / 180.1558 g/mole = 1.665225 x 10^-3
Round to 3 significant figures
1.67 x 10^-3</span>
Answer:
Concentration solution A was 0.5225 M
Explanation:
10.00 mL of solution A was diluted to 50.00 mL and yields 50.00 mL of solution B
According to laws of dilution- ![C_{A}V_{A}=C_{B}V_{B}](https://tex.z-dn.net/?f=C_%7BA%7DV_%7BA%7D%3DC_%7BB%7DV_%7BB%7D)
where,
and
are concentration of solution A and B respectively
and
are volumes of solution A and B respectively
Here
= 0.1045 M,
= 50.00 mL and
= 10.00 mL
Hence, ![C_{A}=\frac{C_{B}V_{B}}{V_{A}}=\frac{(0.1045M\times 50.00mL)}{10.00mL}=0.5225M](https://tex.z-dn.net/?f=C_%7BA%7D%3D%5Cfrac%7BC_%7BB%7DV_%7BB%7D%7D%7BV_%7BA%7D%7D%3D%5Cfrac%7B%280.1045M%5Ctimes%2050.00mL%29%7D%7B10.00mL%7D%3D0.5225M)
So, concentration solution A was 0.5225 M