Answer:
Part a: The rate of the equation for 1st order reaction is given as ![Rate=k[H_2O_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BH_2O_2%5D)
Part b: The integrated Rate Law is given as ![[H_2O_2]=[H_2O_2]_0 e^{-kt}](https://tex.z-dn.net/?f=%5BH_2O_2%5D%3D%5BH_2O_2%5D_0%20e%5E%7B-kt%7D)
Part c: The value of rate constant is 
Part d: Concentration after 4000 s is 0.043 M.
Explanation:
By plotting the relation between the natural log of concentration of 
, the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.
Part a
Rate Law
The rate of the equation for 1st order reaction is given as
Part b
Integrated Rate Law
The integrated Rate Law is given as
Part c
Value of the Rate Constant
Value of the rate constant is given by using the relation between 1st two observations i.e.
t1=0, M1=1.00
t2=120 s , M2=0.91
So k is calculated as
The value of rate constant is
Part d
Concentration after 4000 s is given as
Concentration after 4000 s is 0.043 M.
(a) 485 x 200 mg = 97000 mg of ibuprofen in the bottle
97000 mg x (1g/1000mg) = 97g of ibuprofen in the bottle
97g (1 mol/ 206.5gC13H18O2) = 0.46973 moles of ibuprofen in the bottle
(b) 0.46973 mol C13H18O2 (6.022 x 10^23 molecules/1mol) = 2.8287 x 10^23 molecules of ibuprofen in the bottle
Answer:
The freezing point of the solution is -1.4°C
Explanation:
Freezing point decreases by the addition of a solute to the original solvent, <em>freezing point depression formula is:</em>
ΔT = kf×m×i
<em>Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.</em>
Molality of the solution is:
-moles CaBr₂ (Molar mass:
189.9g ₓ (1mol / 199.89g) = 0.95 moles
Molality is:
0.95 moles CaBr₂ / 3.75kg water = <em>0.253m</em>
Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:
CaBr₂(s) → Ca²⁺ + 2Br⁻
3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.
Replacing:
ΔT = kf×m×i
ΔT = 1.86°C/m×0.253m×3
ΔT = 1.4°C
The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,
<h3>The freezing point of the solution is -1.4°C</h3>
<em />
CO_2
1:2
32gO(1molO/16gO)=2molO
12gC(1molC/12gC)=1molC
