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: 24 moles of 
are produced.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
is accompanied with = 1 mole of
Thus 24 moles of is accompanied with =
of
Thus 24 moles of are produced.
Answer:
Explanation:
(12 x 104 ) x (5 x 10-²) = 6 x 10 ³ 6 x 105 6.0x10²
1. (12 x 104 ) = 1248 or 1.248 x 10^3
2. (1.248 x 10^3)(5 x 10^-2) = 6.240 x 10^1 or 60 rounded to 1 sig fig
I don't understand "= 6 x 10 ³ 6 x 105 6.0x10² "
Weeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
The answer would be D carbon dioxide
