What would be the temperature of 0.6 moles of fluorine that occupy 15 L at 2,300 mmHg

The temperature would be 922.01 °K

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:

P*V=n*R*T

Where P is the gas pressure, V is the volume it occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 2300 mmHgV= 15 Ln= 0.6 molesR= 62.36367  T=?

Replacing:

2300 mmHg* 15 L=0.6 moles*62.36367   *T

Solving:

T= 922.01 °K

The temperature would be 922.01 °K

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na

It idk haha what are we doing?