Answer:
The new temperature is 373 K
Explanation:
Step 1: Data given
Volume air = 5000 mL = 5.0 L
Temperature = 223K
New volume = 8.36 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒V1 = the initial volume = 5.0 L
⇒T1 = the initial temperature = 223 K
⇒V2 = the new volume = 8.36 L
⇒T2 = the new temperature
5.0/223 = 8.36 /T2
T2 = 373 K
The new temperature is 373 K
<h3><u>Answer</u>;</h3>
≈ 4.95 g/L
<h3><u>Explanation;</u></h3>
The molar mass of KCl = 74.5 g/mole
Therefore; 0.140 moles will be equivalent to ;
= 0.140 moles × 74.5 g/mole
= 10.43 g
Concentration in g/L
= mass in g/volume in L
= 10.43/2.1
= 4.9667
<h3> <u> ≈ 4.95 g/L</u></h3>
