The table gives the melting point and boiling point of various elements. Write a compound inequality for each element to show the temperature range of the element in its liquid state. Graph the solutions of each. Suppose you were to set the temperature of each element to its melting point and increase the temperature of each element at the same rate. Which element will remain liquid for the longest amount of time? Which element will reach its boiling point first? Explain.

The answer s below

Step-by-step explanation:

a) The element is going to stay as a liquid if its temperature (t) is within the range: melting point ≤ t < boiling point.

Therefore the inequality for each element is given as:

For Gold:

1064 ≤ t < 2856

For Copper:

1085 ≤ t < 2562

For Iron:

1538 ≤ t < 2861

For lead:

327 ≤ t < 1749

and for Aluminium:

660 ≤ t < 2519

b) To get the element that would be liquid for the longest time, we need to calculate the difference between the melting and boiling points (i.e boiling point - melting point).

For Gold:

Difference = 2856 -  1064 = 1792°

For Copper:

Difference = 2562 -  1064 = 1498°

For Iron:

Difference = 2861 - 1538 = 1323°

For lead:

Difference = 1749 - 327 = 1422°

and for Aluminium:

Difference = 2519 - 660 = 1859°

Since Aluminium has the largest difference between the melting and boiling points, hence it will remain liquid for the longest amount of time.

c) The element with the smallest difference between the melting and boiling points would reach the boiling point first. Therefore iron with a difference of 1323° would reach its boiling point first.

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24