Use the ideal gas law:
PV = nRT
so, T = PV / nR
n=0.5
V= 120 dm^3 = 120 L (1 dm^3 = 1 L)
R = 1/12
P = 15,000 Pa = 0.147 atm (1 pa = 9.86 10^{-6} )
Put the values:
T = PV / nR
T = (0.147) (120) / (0.5) (1/12)
T= 426 K
Answer:
<u>= 2.2 g pf S. produced</u>
Explanation:
Balanced Reaction equation:
→ 
1 mole of H2S - 34.1g
? moles - 3.2g
= 3.2/34.1 =<u> 0.09 moles of H2S</u>
Also,
1 mole of S02 - 64.07 g
? moles - 4.42g
= 4.42/64.07 <u>= 0.069 moles of SO2</u>
<u />
<em>Meaning SO2 is the limiting reagent</em>
Finally, 3 moles of S - 32g of sulphur
0.069 mole = ? g of Sulphur
= 0.069 x 32
<u>= 2.2 g pf S.</u>
Preparing 15 mg/gl working standard solution from a 20 mg/dl stock solution will require the application of the dilution principle.
Recalling the principle:
initial volume x initial molarity = final volume x final molarity
Since we were not given any volume to work with, we can as well just take an arbitrary volume to be prepared. Let's assume that the stock solution is 10 mL and we want to prepare 15 mg/gl from it:
Applying the dilution principle:
10 x 20 = final volume x 15
final volume = 200/15
= 13.33 mL
This means that in order to prepare 13.33 mL, 15 mg/l working standard solution from 10 ml, 20 mg/dl stock solution, 3.33 mL of the diluent must be added to the stock solution.
More on dilution principle can be found here: brainly.com/question/11493179
