Answer:
When barium chloride (BaCl 2) is dissolved in water, the water conducts electricity. In what form will the dissolved BaCl 2 be found? a. as Ba 2+ and Cl - ions b. as Ba atoms and Cl 2 molecules
Explanation:
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
<h3>
Answer:</h3>
812 kPa
<h3>
Explanation:</h3>
- According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
- Mathematically,

At varying pressure and volume;
P1V1=P2V2
In this case;
Initial volume, V1 = 2.0 L
Initial pressure, P1 = 101.5 kPa
Final volume, V1 = 0.25 L
We are required to determine the new pressure;

Replacing the known variables with the values;

= 812 kPa
Thus, the pressure of air inside the balloon after squeezing is 812 kPa
A MOLECULE IS MADE OF TWO OR MORE ELEMENTS CHEMICALLY COMBINED IS KNOWN AS A COMPUND.
A MOLECULE IS MADE OF TWO ATOMS IS JUST AN ELEMENT.
A MOLECULE MADE OF TWO OR MORE ELEMENTS IS KNOWN AS A COMPUND.
Answer:
(a) The rate of formation of K2O is 0.12 M/s.
The rate of formation of N2 is also 0.12 M/s
(b) The rate of decomposition of KNO3 is 0.24 M/s
Explanation:
(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.
Rate of formation of O2 is 0.3 M/s
Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s
Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.
Rate of formation of N2 = (2×0.3/5) = 0.12 M/s
(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.
Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s