Iron-59 which is used to diagnose anaemia has a half life of 45days, what fraction of it is left in 90 days

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

Ci: initial concentration
Vi: initial volume
Cf: final concentration
Vf: final volume

<h3>Molarity of the solution in this case</h3>

In this case, you know:

Ci= 4 M
Vi= 7 mL
Cf= ?
Vf= 30 mL

Replacing in the definition of dilution:

4 M× 7 mL= Cf× 30 mL

Solving:

(4 M× 7 mL)÷ 30 mL= Cf

In summary, the molarity of a solution prepared by diluting 7.0 mL of 4.0 M solution to a volume of 30 mL is 0.93 mL.

Learn more about dilution:

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8 0

2 years ago

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g )

Aleksandr [31]

Answer:

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

$rate = k*[O_{3}][NO]$ (2)

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

$rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s$

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

$\frac{\Delta[NO_{2}]}{\Delta t} = rate$

$\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}$

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!

5 0

3 years ago