Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the substance remaining after time , and is a constant that is characteristic of the substance. for the radioactive isotope iron-55, is if the original amount of iron-55 in a sample is 32.2 mg, how much iron-55 remains after 2.41 years have passed? mg

Iron remains = 17.49 mg

Explanation:

Half life of iron -55 = 2.737 years (Source)

Where, k is rate constant

So,  

The rate constant, k = 0.2533 year⁻¹

Time = 2.41 years

= 32.2 mg

Using integrated rate law for first order kinetics as:

Where,  

is the concentration at time t

is the initial concentration

So,  

Iron remains = 17.49 mg

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol