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.
The answer is A.
Which is... KCI
Answer:

Explanation:
Molarity is found by dividing the moles of solute by liters of solution.

We know the molarity is 1.2 M (mol\liter) and there are 2.3 liters of solution. Substitute the known values into the formula.

Since we are solving for x, we must isolate the variable. It is being divided by 2.3 and the inverse of division is multiplication. Multiply both sides by 2.3 liters.

In a solution with a molarity of 1.2 and 2.3 liters of solution, there are 2.76 moles.
Answer:
P_2 =0.51 atm
Explanation:
Given that:
Volume (V1) = 2.50 L
Temperature (T1) = 298 K
Volume (V2) = 4.50 L
at standard temperature and pressure;
Pressure (P1) = 1 atm
Temperature (T2) = 273 K
Pressure P2 = ??
Using combined gas law:




Answer:
a. 0.5 mol
b. 1.5 mol
c. 0.67
Explanation:
Fe3+ + SCN- -----> [FeSCN]2+
a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium
b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium
c. KC = 0.5/(0.5*1.5) = 0.67