Answer:
15. 2.66 moles .
16. 2.09L.
Explanation:
Molarity of a solution is simply defined as the mole of solute per unit litre of the solvent. Mathematically, it is represented as:
Molarity = mole /Volume.
With the above formula, let us answer the questions given above
15. Data obtained from the question include the following:
Volume of solution = 1.4L
Molarity = 1.9M
Mole of solute =.?
Molarity = mole /Volume
1.9 = mole / 1.4
Cross multiply
Mole = 1.9 x 1.4
Mole = 2.66 moles
Therefore, the mole of the solute present in the solution is 2.66 moles.
16. Data obtained from the question include the following:
Mole of solute = 0.46 mole
Molarity = 0.22M
Volume of solvent (water) =.?
Molarity = mole /Volume
0.22 = 0.46/Volume
Cross multiply
0.22 x Volume = 0.46
Divide both side 0.22
Volume = 0.46/0.22
Volume = 2.09L
Therefore, 2.09L of water is required.
5. 1.16 x 
moles moles of pennies would be required to equal the mass of the moon.
6. 12.86 moles of ethanol are in a 750 ml bottle of vodka.
Explanation:
5 .Data given:
mass of penny = 2.5 grams
atomic mass of penny = 62.93 grams/mole
moles present in mass of the moon given as = 7.3 x 
kg
number of moles = 
number of moles = 
0.039 moles of penny is present in 2.5 grams
0.039 moles of penny in 2.5 grams of it
so, x moles in 7.3 X 
grams
x = 1.16 x moles
so when the mass of the penny given is equal to the mass of moon, number of moles of penny present is 1.1 x .
6.
Given:
vodka = 40% ethanol
volume of vodka bottle = 750 ml
moles of ethanol =?
density of ethanol =0.79 g/ml
atomic mass of ethanol = 46.07 grams/mole
so, from the density of ethanol given we can calculate how much ethanol is present in the solution.
density = 
density x volume = mass
0.79 x 750 = 592.5 grams
number of moles = 
number of moles of ethanol = 
= 12.86 moles of ethanol
Explanation:
If 50.0 grams of Zinc are reacted with 50.0 grams of Hydrogen Chloride ... 50.09 Zn x 1 mol Zn , Imol ZnCl2 , 136.4g. ... If a decomposition reaction produces a 75.0% yield for the oxygen by mass (128.0 grams were.
Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ
