Answer:
evaporated
Explanation:
Once the solution is evaporated there will only be salt left, since the only other thing in the soulution is water.
Answer:
Part A
K = (K₂)²
K = (K₃)⁻²
Part B
K = √(Ka/Kb)
Explanation:
Part A
The parent reaction is
2Al(s) + 3Br₂(l) ⇌ 2AlBr₃(s)
The equilibrium constant is given as
K = [AlBr₃]²/[Al]²[Br₂]³
2) Al(s) + (3/2) Br₂(l) ⇌ AlBr₃(s)
K₂ = [AlBr₃]/[Al][Br₂]¹•⁵
It is evident that
K = (K₂)²
3) AlBr₃(s) ⇌ Al(s) + 3/2 Br₂(l)
K₃ = [Al][Br₂]¹•⁵/[AlBr₃]
K = (K₃)⁻²
Part B
Parent reaction
S(s) + O₂(g) ⇌ SO₂(g)
K = [SO₂]/[S][O₂]
a) 2S(s) + 3O₂(g) ⇌ 2SO₃(g)
Ka = [SO₃]²/[S]²[O₂]³
[SO₃]² = Ka × [S]²[O₂]³
b) 2SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
Kb = [SO₃]²/[SO₂]²[O₂]
[SO₃]² = Kb × [SO₂]²[O₂]
[SO₃]² = [SO₃]²
Hence,
Ka × [S]²[O₂]³ = Kb × [SO₂]²[O₂]
(Ka/Kb) = [SO₂]²[O₂]/[S]²[O₂]³
(Ka/Kb) = [SO₂]²/[S]²[O₂]²
(Ka/Kb) = {[SO₂]/[S][O₂]}²
Recall
K = [SO₂]/[S][O₂]
Hence,
(Ka/Kb) = K²
K = √(Ka/Kb)
Hope this Helps!!!
<h3>The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)</h3><h3>Further explanation </h3>
The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released
Q in = Q out
Heat can be calculated using the formula:
Q = mc∆T
Heat released by anthracene= Heat absorbed by water
Heat absorbed by water =
mol of anthracene (MW=178,23 g/mol)
The enthalpy of combustion per mole of anthracene :
<span>15.4 milligrams
The ideal gas law is
PV = nRT
where
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = Ideal gas constant (8.3144598 L*kPa/(K*mol) )
T = absolute temperature.
So let's determine how many moles of gas has been collected.
Converting temperature from C to K
273.15 + 25 = 298.15 K
Converting pressure from mmHg to kPa
753 mmHg * 0.133322387415 kPa/mmHg = 100.3917577 kPa
Taking idea gas equation and solving for n
PV = nRT
PV/RT = n
n = PV/RT
Substituting known values
n = PV/RT
n = (100.3917577 kPa 0.195 L) / (8.3144598 L*kPa/(K*mol) 298.15 K)
n = (19.57639275 L*kPa) / (2478.956189 L*kPa/(mol) )
n = 0.007897031 mol
So we have a total of 0.007897031 moles of gas particles.
Now let's get rid of that percentage that's water vapor. The percentage of water vapor is the vapor pressure of water divided by the total pressure. So
24/753 = 0.03187251
The portion of hydrogen is 1 minus the portion of water vapor. So
1 - 0.03187251 = 0.96812749
So the number of moles of hydrogen is
0.96812749 * 0.007897031 mol = 0.007645332 mol
Now just multiple the number of moles by the molar mass of hydrogen gas. Start with the atomic weight.
Atomic weight hydrogen = 1.00794
Molar mass H2 = 1.00794 * 2 = 2.01588 g/mol
Mass H2 = 2.01588 g/mol * 0.007645332 mol = 0.015412073 g
Rounding to 3 significant figures gives 0.0154 g = 15.4 mg</span>
