Answer:
1 mole of gas = 22.4L
2.5 moles of gas takes up = ( 22.4 L/ 1 mole ) x 2.5 mole
= 56 L
Explanation:
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Its going to be 2.81 x 1023 atoms
<span>the behavior of the above pair of substances</span> is soluble
Volume of Argon V1 = 5.0 L
Pressure of Argon P1 = 2 atm
Final temperature T2 = 30 C = 30 + 273 = 303 K
Volume at final temperature V2= 6 L
Pressure at final temperature P2 = 8 atm
We know that (P1 x V1) / T1 = (P2 x V2) / T2
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48
Initial Temperature T1 = 3030 / 48 = 63.12
Initial Temperature = -209. 8 C
Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:

Now, find moles of
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M