Answer:
1. [H+] = 1 x 10^-4mol/L
2. [OH-] = 1 x 10^-10mol/L
Explanation:
1. From the question given, the pH is 4.0. We can obtain the value of the value of the hydrogen ion concentration [H+] as follows:
pH = —Log[H+]
4 = —Log[H+]
—4 = Log[H+]
Anti-log of —4
[H+] = 1 x 10^-4mol/L
2. Recall: [H+] x [OH-] = 1 x 10^-14
But [H+] = 1 x 10^-4mol/L
1 x 10^-4 x [OH-] = 1 x 10^-14
Divide both side by 1 x 10^-4mol/L
[OH-] = 1 x 10^-14/1 x 10^-4
[OH-] = 1 x 10^-10mol/L
Given:
K = 0.71 = Kp
The reaction of sulphur with oxygen is
S(s) + O2(g) ---> SO2(g)
initial Pressure 6.90 0
Change -x +x
Equilibrium 6.90-x x
Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)
4.899 - 0.71x = x
4.899 = 1.71x
x = 2.86 atm = pressure of SO2 formed
temperature = 950 C = 950 + 273.15 K = 1223.15 K
Volume = 50 L
Let us calculate moles of SO2 formed using ideal gas equation as
PV = nRT
R = gas constant = 0.0821 L atm / mol K
putting other values
n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles
Moles of Sulphur required = 1.42 moles
Mass of sulphur required or consumed = moles X atomic mass of sulphur
mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur
CH4 + 2O2 --> CO2 + 2H2O
On the reactant side you start with 1 carbon, 4 hydrogen, and 2 oxygen. On the product side you start with 1 carbon, 2 hydrogen, and 3 oxygen. In order to get them equal, you need to put 2 in front of the H2O which equals out the number of Hydrogen on both sides. But Now we must balance the Oxygens. Because of the H2O we now have 4 Oxygens on the product side and only 2 on the reactant. In order to balance this, we put a 2 in front of the O2 giving us 4 hydrogen on both sides, Balancing the equation
Answer:
Answer. 600 g CaCl2 would be required to make 2 L of a 3.5 M solution.
