Answer:
Therefore it takes 8.0 mins for it to decrease to 0.085 M
Explanation:
First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.
A→ product
Let the concentration of A = [A]
![\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]](https://tex.z-dn.net/?f=%5Ctextrm%7Brate%20of%20reaction%7D%3D-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dk%5BA%5D)
![k=\frac{2.303}{t} log\frac{[A_0]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%20log%5Cfrac%7B%5BA_0%5D%7D%7B%5BA%5D%7D)
[A₀] = initial concentration
[A]= final concentration
t= time
k= rate constant
Half life: Half life is time to reduce the concentration of reactant of its half.

Here 


To find the time takes for it to decrease to 0.085 we use the below equation
![k=\frac{2.303}{t} log\frac{[A_0]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%20log%5Cfrac%7B%5BA_0%5D%7D%7B%5BA%5D%7D)
![\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}](https://tex.z-dn.net/?f=%5CRightarrow%20t%3D%5Cfrac%7B2.303%7D%7Bk%7D%20log%5Cfrac%7B%5BA_0%5D%7D%7B%5BA%5D%7D)
Here ,
, [A₀] = 0.13 m and [ A] = 0.085 M


Therefore it takes 8.0 mins for it to decrease to 0.085 M
because iodine is an atom and it can permeate the pores of the plastic bag.
Answer:
1132.8 ml of water
Explanation:
you have an aqueous solution contains 158.2 g KOH per liter
so concentration =158.2/56 = 2.825M
Molarity =2.825
that means you have 2.825 moles of KOH in 1.00L solution
Mass of Soluet(KOH)= 152.8g
Volume of solution= 1.00L
density of solution= 1.13g/cm3 =1.13g/ml
therefore mass of solution = VolumeX density = 1000mL X 1.13g/ml.=1130g
Mass of solvent(water)= mass of solution- mass of solute(KOH)=1130-152.8= 997.2g
Molality= moles of solute/mass of solvent(Kg)
=2.825/(997.2/1000)= 2.832molal
to prepare a 0.250 molal solution of KOH, starting with 100.0ml ofthe orginal solution
0.250*X =2.832 *100
X = 1132.8 ml of water you have to add
Answer:
Mole fraction for solute = 0.1, or 10%
Molality = 6.24 mol/kg
Explanation:
22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH
Mass of solution = 100 g
Mass of solute = 22.3 g
Mass of solvent = 100 g - 22.3g = 77.7 g
Let's convert the mass to moles
22.3 g . 1mol/ 46 g = 0.485 moles
77.7 g. 1mol / 18 g = 4.32 moles
Total moles = 4.32 moles + 0.485 moles = 4.805 moles
Xm for solute = 0.485 / 4.805 = 0.100 → 10%
Molality → mol/ kg → we convert the mass of solvent to kg
77.7 g. 1 kg / 1000g = 0.0777 kg
0.485 mol / 0.0777 kg = 6.24 m