1. c. 35
2. <span>d. 1.26 x 10^24 molecules
3. </span><span>d. 303.6 g</span>
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 is: there are two compounds with molecular formula C₂H₆O, one alcohol and one ethar.
Alcohol with molecular formula C₂H₆O is ethanol or ethyl alcohol (C₂H₅OH).
Ethanol is a volatile, flammable, colorless liquid with characteristic odor.
Ether with molecular formula C₂H₆O is dimethyl ether (CH₃OCH₃).
Dimethyl ether is a colorless gas.
When writing an ionic compound formula, a "molecular" form is used. The formula is made with allowance for ion charges.
For example,
Ca²⁺ and NO₃⁻ ⇒ Ca(NO₃)₂
Al³⁺ and SO₄²⁻ ⇒ Al₂(SO₄)₃
