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:
= 61.25 g
= 88.75 g
Explanation:
=
= 50 g
⇒
=
= 1.25 (moles)
2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O
2 : 1 : 1 : 2
1.25 (moles)
⇒
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒
= 0.625 × 98 = 61.25 g
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒
= 0.625 × 142 = 88.75 g
Answer:
1,31÷2 =10,11
Explanation:
c10h22+31÷2o2=10co2+11h2o
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