The volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.
<h3>How to calculate volume?</h3>
The concentration of a solution can be calculated using the following formula:
C1V1 = C2V2
Where;
- C1 = initial concentration = 1.5M
- C2 = final concentration = 0.54M
- V1 = initial volume = ?
- V2 = final volume = 0.18L
1.5 × V1 = 0.54 × 0.18
1.5V1 = 0.0972
V1 = 0.0972 ÷ 1.5
V1 = 0.065L
Therefore, the volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.
Learn more about volume at: brainly.com/question/1578538
When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
Answer:
Gasification refers to the conversion of coal to a mixture of gases, including carbon monoxide, hydrogen, methane, and other hydrocarbons, depending on the conditions involved.
It will also be halved... but the relationship is with Kelvin not c. So 273 +273 = 546 K. half of that is 273 K, then subtract 273... you get 0 degrees c
Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
