On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
Answer 8.0 L.
2.0L / 5.0 moles = x / 20.0 => x = 20 / 5 * 2 = 8
Given :
A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.
To Find :
The volume of the gas after it is heated.
Solution :
Since, their is no information about pressure in the question statement let us assume that pressure is constant.
Now, we know by ideal gas equation at constant pressure :
Hence, this is the required solution.
Answer:
0.14 M
Explanation:
To determinate the concentration of a new solution, we can use the equation below:
C1xV1 = C2xV2
Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL
1.50x55.0 = C2x278
C2 = 0.30 M
The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL
Then,
0.30x139 = C2x294
C2 = 0.14 M
