<span>The temperature in the tire increased, causing an increased tire pressure. :D</span>
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
V1M1 = V2M2
<span>V1 × 2.5 = 1 × 0.75,
so V1 = 0.75/2.5
= 0.3 </span>
To
determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample
of the compound with the given composition. Then, we calculate for the number
of moles of each element. We do as follows:<span>
mass moles
C 56.79 4.73
H 6.56 6.50
O 28.37 1.77
N 8.28 0.59
Dividing the number of moles of each element with
the smallest value, we will have the empirical formula:
</span> moles ratio
C 4.73 / 0.59 8
H 6.50 / 0.59 11
O 1.77 / 0.59 3
N 0.59 / 0.59 1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>
Answer:
1.387 moles
Explanation:
Step 1:
The balanced equation for the reaction. This is illustrated below:
4Fe + 3O2 —> 2Fe2O3
Step 2:
Determination of the number of mole of Fe in 155.321g of Fe. This can be achieved by doing the following:
Mass of Fe = 155.321g
Molar Mass of Fe = 56g/mol
Number of mole of Fe =?
Number of mole = Mass/Molar Mass
Number of mole of Fe = 155.321/56
Number of mole of Fe = 2.774 mol
Step 3:
Determination of the number of mole of rust (Fe2O3) produced. This is illustrated below:
From the balanced equation above,
4 moles of Fe produced 2 moles of Fe2O3.
Therefore, 2.774 moles of Fe will produce = (2.774 x 2)/4 = 1.387 moles of Fe2O3.
Therefore, 1.387 moles of rust (Fe2O3) is produced from the reaction
