Answer:
The Ka is 9.11 *10^-8
Explanation:
<u>Step 1: </u>Data given
Moles of HX = 0.365
Volume of the solution = 835.0 mL = 0.835 L
pH of the solution = 3.70
<u>Step 2:</u> Calculate molarity of HX
Molarity HX = moles HX / volume solution
Molarity HX = 0.365 mol / 0.835 L
Molarity HX = 0.437 M
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<u>Step 3:</u> ICE-chart
[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4
Initial concentration of HX = 0.437 M
Initial concentration of X- and H3O+ = 0M
Since the mole ratio is 1:1; there will react x M
The concentration at the equilibrium is:
[HX] = (0.437 - x)M
[X-] = x M
[H3O+] = 1.995*10^-4 M
Since 0+x = 1.995*10^-4 ⇒ x=1.995*10^-4
[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M
[X-] = x = 1.995*10^-4 M
<u>Step 4: </u>Calculate Ka
Ka = [X-]*[H3O+] / [HX]
Ka = ((1.995*10^-4)²)/ 0.437
Ka = 9.11 *10^-8
The Ka is 9.11 *10^-8
wrong. If its for the study island its potential energy only.
Answer: When a fire is burning, chemical energy is turned into visible energy in the form of light and thermal energy in the form of heat.
Explanation:
Explanation and answer:
The molar volume of a gas is 22.4 L at 0 deg. C.
So the molar mass of the gas is, by proportion,
0.16 g * (22400 mL)/(240 mL) = 14.93 g
The molar mass of (CH4)n = 12+4(1) = 16.
So n = 15.93/16 = 1, or the molecular formula is CH4.
Note: The temperature at which the volume was observed was not given. If 240 cm^3 was observed at 20 deg.C, then the volume at 0 deg.C would be
V=240*(273+0)/(273+20) = 223.6
The molar mass = 0.16*22400/(223.6) = 16.03
which gives n = 16/16.03 = 1 again, but more accurately.
Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.
