Answer:
Mass in kg = 4.7*10^19 kg
Mass in tons = 5.2*10^16 tons
Explanation:
<u>Given:</u>
Total volume of sea water = 1.5*10^21 L
Mass % NaCl in seawater = 3.1%
Density of seawater = 1.03 g/ml
<u>To determine:</u>
Total mass of NaCl in kg and in tons
<u>Calculation:</u>
Unit conversion:
1 L = 1000 ml
The volume of seawater in ml is:
To convert mass from g to Kg:
1000 g = 1 kg
To convert mass from g to tons:
1 ton = 9.072*10^6 g
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K =
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M