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See detailed solution below

Explanation:

Now, let us solve each question step by step.

1) Given that;

N = amount of radioactive material at time t

No= amount of radioactive material originally present

t = time required for N amount of radioactive material to remain

t1/2= half life of the radioactive material

So,

N/No = (1/2)^t/t1/2

N/60 = (1/2)^22.8/3.8

N/60= (1/2)^6

N/60=1/64

64 N = 60

N = 60/64

N = 0.9 g

2) N/No = (1/2)^t/t1/2

N/8 = (1/2)^103.2/34.4

N/8 = (1/2)^3

N/8 = 1/8

8N = 8

N = 1 L

3)  N/No = (1/2)^t/t1/2

50/No =  (1/2)^18/6

50/No =  (1/2)^3

50/No = 1/8

No = 50 * 8

No = 400 micrograms

4)  N/No = (1/2)^t/t1/2

62.5/500 =  (1/2)^26/t1/2

0.125 = (1/2)^26/t1/2

Since 0.125 = 1/8

1/8 =  (1/2)^26/t1/2

(1/2)^3 = (1/2)^26/t1/2

Hence;

3 = 26/t1/2

t1/2 = 26/3

t1/2 = 8.7 hours

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