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:
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Explanation:
Answer:12 mol
Explanation: both vessels are at the same temp and pressure (and the pressure is low and/or the temperature high).
6.7mol per 1.3L = 6.7/1.3 mol/L
so in 2.33L = 6.7*2.33/1.3 = 12 mol
Answer:
Condensation 212F or 100C, Freezing 32F or 0C
Explanation:
Condensation 212 degrees Fahrenheit or 100 degrees Celsius.
Freezing point 32 degrees Fahrenheit or 0 degrees Celsius.
Metal period- Reactivity Gets smaller as you move from left to right.
Group- Reactivity gets bigger as you move down a group.
