Answer:
The value of the equilibrium constant KC is 1.244
Explanation:
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C
At equilibrium, [A] = 0.430 M and [C] = 0.720 M. Calculate the value of the equilibrium constant, Kc
Step 1: The balanced equation
A+2B<->C
Step 2: The initial concentrations
[A] = 0.550 M
[B]= 1.40 M
[C] = 0.600 M
Step 3: The concentraions at equilibrium
[A] = 0.550 -X = 0.430 M
[B]= 1.40 -2X M
[C] = 0.600 + X = 0.720 M
X = 0.120 M
[A] = 0.550 - 0.120 = 0.430 M
[B]= 1.40 -2*0.120 = 1.16 M
[C] = 0.600 + 0.120 = 0.720 M
Step 4: Calculate Kc
Kc = [C] / [A][B]²
Kc = 0.720 / (0.430*1.16²)
Kc = 1.244
The value of the equilibrium constant KC is 1.244
The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .
The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.
The thermochemical equation for the combustion of benzene is;
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol
We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.
Learn more: brainly.com/question/13164491
Hello!
We know that by the Law of Avogrado, for each mole of substance we have 6.02 * 10²³ atoms, if:
The molar mass of water (H2O)
H = 2 * (1u) = 2u
O = 1 * (16u) = 16u
---------------------------
The molar mass of H2O = 2 + 16 = 18 g / mol
If:
1 mol we have 6.02 * 10²³ atoms
1 mole of H2O we have 18 g
Then we have:
18 g ------------- 6.02 * 10²³ atoms
5 g -------------- x
I Hope this helps, greetings ... DexteR! =)
Yes, the atomic radius increases as you move down a group of elements.
this is true
going down leads to valence electrons that are further away from nucleus -> less electrostatic attraction -> less pull towards nuc. -> greater radius/volume taken
Methane is the compound CH4, and burning it uses the reaction:
CH4 + O2 -> CO2 + H2O, which is rather exothermic. To find the heat released by burning a certain amount of the substance, you should look at the bond enthalpy of each compound, and then compare the values before and after the reaction. In methane, there are 4 C-H bonds, which have bond energy of 416 kj/mol, resulting in a total bond energy of 1664 kj/mol. O2 is 494 kj/mol. Therefore we have a total of 2080 kj/mol on the left side. On the right side we have CO2, which has 2 C=O bonds, each at 799 kj/mol each, resulting in 1598 kj/mol, and H2O has 2 O-H bonds, at 459kj/mol each, resulting in a total of 2516 kj/mol on the right hand side. Now, this may be confusing because the left hand side seems to have less heat than the right, but you just need to remember: making minus breaking, which results in a total change of 436kj/mol heat evolved.
Now it is a simple matter of find the mols of CH4 reacted, using n=m/mr.
n = 9.5/16.042 = 0.592195 mol
Therefore, if we reacted 0.592195 mol, and we produced 436 kj for one mol, the total amount of energy evolved was 436*<span>0.592195 kj, or 258.197 kj.</span>
