A 0.25g sample of Lead-210 has a half life of 20.4 years. Will all of the Lead undergo radioactive decay in 150 years? Explain. (I really need help on this this the last question for my assignment but i’m completely lost since i don’t think she covered how we’re supposed to solve this and explain it)

Using the equation A(t) = 400e-.032t

a) replace t with 4 so A(4) = 400e((-.032)(4))

The hardest part about this is making sure to use order of operations. Be certain it works like this:

A(4) = 400e-.128

A(4) = 400(.8799)

A(4) = 351.9 grams

b) A(8) = 400e((-.032)(8)) = 309.7 grams

c) A(20) = 400e((-.032)(20)) = 210.9 grams

Note here that even after 20 years, not quite half of the original amount is gone. So, we can anticipate that in finding the half life, that our answer should be slightly greater than 20 years.

d) 200 = 400e(-.032t)

Divide both sides of the equation by 400.

.5 = e(-.032t)

Change this to logarithmic form.

Ln .5 = -.032t

-.6931≈ -.032t

t ≈ 21.7 years

Explanation:

sure what is it? ^w^

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