Answer:
134.8 seconds is the half-life (in seconds) of the reaction for the initial 
concentration
Explanation:
Half life for second order kinetics is given by:
Integrated rate law for second order kinetics is given by:
= half life
k = rate constant
= initial concentration
a = Final concentration of reactant after time t
We have :
Initial concentration of ![C_2F_4=[a_o]=\frac{0.438 mol}{2.42 L}=0.1810 mol/L](https://tex.z-dn.net/?f=C_2F_4%3D%5Ba_o%5D%3D%5Cfrac%7B0.438%20mol%7D%7B2.42%20L%7D%3D0.1810%20mol%2FL)
Rate constant = k = 
134.8 seconds is the half-life (in seconds) of the reaction for the initial concentration
Answer:could it be kilo
Explanation:it looks like it has the least letters
For balancing acidic solutions, we would need to add H+ ions to the correct side of the equation to balance the total number of atoms and the overall charge.
The system will then shift to the left, making more reactants.
Answer:
T° freezing solution → -11.3°C
T° boiling solution → 103.1 °C
Explanation:
Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"
This salt dissociates as this:
SnCl₄ (aq) → 1Sn⁴⁺ (aq) + 4Cl⁻ (aq) (so i =5)
The formula for the colligative property of freezing point depression and boiling point elevation are:
ΔT = Kf . m . i
where ΔT = T° freezing pure solvent - T° freezing solution
ΔT = Kb . m . i
where ΔT = T° boiling solution - T° boiling pure solvent
Freezing point depression:
0° - T° freezing solution = 1.86°C/m . 1.22 m . 5
T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C
Boiling point elevation:
T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5
T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C
