Explanation:
Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. The resulting solution is called a saturated solution.
Answer:
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 4 g
volume = 2 cm³
We have
We have the final answer as
Hope this helps you
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
Answer:
Explanation:
Hexanoic acid [CH 3(CH 2) 4COOH] is barely soluble in water (about 1.0 g/100 g of water). Palmitic acid [CH 3(CH 2) 14COOH], with its large nonpolar hydrocarbon component, is essentially insoluble in water. The carboxylic acids generally are soluble in such organic solvents as ethanol, toluene, and diethyl ether.
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Answer:
164 g
Explanation:
Part A
If the pH of a 0.133 M solution is 10.66, what is the ionization constant of the base?
Express your answer using two significant figures.
Answer:
Kb = 1.6*10^-6
Explanation:
The given reaction is:
The ionization constant of the base Kb is given as:
The pH = 10.66
therefore, pOH = 14-pH = 14-10.66 =3.34
[N2H5+] = [OH-] = 4.57*10^-4M
[N2H4] = 0.133 M
Based on eq(1)
Answer:
(a) The rate constant is 3.61×10^-3 s^-1
(b) 7.12×10^4 s
Explanation:
(a) Log (K2/K1) = Ea/2.303R × [1/T1 - 1/T2]
K1 = 8.33×10^-6 s^-1
Ea = 245 kJ = 245,000 J
R = 8.314 J/mol.K
T1 = 427°C = 427+273 = 700 K
T2 = 545°C = 546+273 = 818 K
Log (K2/8.33×10^-6) = 245,000/2.303 × [1/700 - 1/818]
Log (K2/8.33×10^-6) = 2.637
K2/8.33×10^-6 = 10^2.637
K2 = 8.33×10^-6 × 433.51 = 3.61×10^-3 s^-1
(b) The relationship between temperature and the time required for reactants to be consumed is inverse
t2 = T1t1/T2
T1 = 427 °C = 700 K
t1 = 8.32×10^4 s
T2 = 545 °C = 818 K
t2 = 700×8.32×10^4/818 = 7.12×10^4 s