P1V1 = P2V2
P1 = 720 mmHg
V1 = 450. mL
P2 = 760 mmHg (this is the pressure at STP)
Use these to solve for V2:
(720)(450) = 760V2
V2 = 426 mL
Answer:
Products
Explanation:
In a chemical reaction, the atoms and molecules produced by the reaction are called products
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
Answer:
Mg S2 O3
Explanation:
.691 g of Mg is .284 mole
1.84 g of S is .5739 mole
1.365 g of O is .8531 mole you can see the ratio is ~ 1 :2 :3
Mg S2 O3
During a phase change the temperature does not change since all of the heat is being absorbed in order to break the intermolecular forces. Due to that, the formula will not need to have T in it and is actually q=nΔH(v).
n=the number of moles (in this case 2.778mol of water since you divide 50g by 18g/mol).
ΔH(v)=the molar heat of vaporization (in this case 40.7kJ/mol).
q=the heat that must be absorbed
q=2.778mol×40.7kJ/mol
q=113.1kJ
Therefore the water needs to absorb 1.13×10²kJ.
I hope this helps. Let me know if anything is unclear.
