The radioactive substance cesium-137 has a half-life of 30 years. the amount at (in grams) of a sample of cesium-137 remaining after t years is given by the following exponential function.
at(t)=458(1/2)^(t/30)
find the amount of the sample remaining after 20 years and after 50 years. round your answers to the nearest gram as necessary.

The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

Explanation:

We are given a function that calculates the amount of sample remaining after 't' years, which is:

Putting values in above equation:

Hence, the amount of sample left after 20 years is 288.522 g

Putting values in above equation:

Hence, the amount of sample left after 50 years is 144.26 g

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

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