Answer: A) 0.20 L
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
According to the neutralization law,
where,
= molarity of stock solution = 10.0 M
= volume of stock solution = ?
= molarity of dilute solution = 0.50 M
= volume of dilute solution = 4.0 L
Therefore, the volume of 10.0 M solution needed to prepare 4.0 L of 0.50 M is 0.20 L
Given:
Moles of H2 = 0.300
Moles of I2 = 0.400
Moles of HI = 0.200
Keq = 870
To determine:
Amounts of the mixture at equilibrium
Explanation:
H2(g) + I2(g) ↔ 2HI(g)
Initial 0.3 0.4 0.2
Change -x -x +2x
Eq (0.3-x) (0.4-x) (0.2+2x)
Keq = [HI]²/[H2][I2]
870 = (0.2+2x)²/(0.3-x)(0.4-x)
x = 0.29 moles
Amounts at equilibrium:
[HI] = 0.2 + 2(0.29) = 0.78 moles
[H2] = 0.3-0.29 = 0.01 moles
[I2] = 0.4-0.29 = 0.11 moles
Exothermic processes Endothermic processes
making ice cubes melting ice cubes
formation of snow in clouds conversion of frost to water vapor
condensation of rain from water vapor evaporation of water
Assuming ideality:
PV=nRT
Pv= (mass/molarmass)RT
Solve for the mass :-)
Answer:
Yes.
Explanation:
Yes, this method is accurate just as the slow titration because in both type of titrations we want to measure the point at which the indicator change the colour of the solution. Both fast and slow titrations are the same in their function i.e. both tell us the point or the amount at which the indicator change the colour of the solution so we can conclude from this discussion that fast titration gives the same result just as the slow titration.