<span>Calculating the moles and the moles ratio of the elements gives us the ratio of atoms in each element.
Converting the percentage of element into grams
40.25% carbon = 40.25/100 = .4025 * 100 g of carbon = 40.25g of C
6.19% hydrogen = 6.19/100 = .0619 * 100g g of hydrogen = 6.19g of H
8.94% oxygen = 8.94/100 = .0819 * 100 g of oxygen = 8.19g of O
44.62% bromine = 44.62/100 = .4462 * 100 g of bromine = 44.62g of Br
Converting the grams of element into moles
(48.38 g C) (1 mol/ 12.10 g C) = 4.028 mol C
(8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H
(53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O
(44.62g of Br)(0.012515018021626 moles) = 0.55842 mol Br
Calculating the moles ratio of elements by dividing the small number of moles of an element
4.028 mol C /0.55842 = 7.2 mol C x 5 = 36 mol C
8.056 mol H / 0.55842 = 14.42 mol H = 72 mol H
3.336 mol O / 0.55842 = 5.97 mol O = 30 mol O
0.55842 mol Br / 0.55842 = 1mol Br = 5 mol Br
So the empirical formula is (C6H12O5)6Br5</span>
Answer: Elastic
Explanation: Both Objects had there momentum and kinetic energy conserved.
Answer:
a. 7278 K
b. 4.542 × 10⁻³¹
Explanation:
a.
Let´s consider the following reaction.
N₂(g) + O₂(g) ⇄ 2 NO(g)
The reaction is spontaneous when:
ΔG° < 0 [1]
Let's consider a second relation:
ΔG° = ΔH° - T × ΔS° [2]
Combining [1] and [2],
ΔH° - T × ΔS° < 0
ΔH° < T × ΔS°
T > ΔH°/ΔS°
T > (180.5 × 10³ J/mol)/(24.80 J/mol.K)
T > 7278 K
b.
First, we will calculate ΔG° at 25°C + 273.15 = 298 K
ΔG° = ΔH° - T × ΔS°
ΔG° = 180.5 kJ/mol - 298 K × 24.80 × 10⁻³ kJ/mol.K
ΔG° = 173.1 kJ/mol
We can calculate the equilibrium constant using the following expression.
ΔG° = - R × T × lnK
lnK = - ΔG° / R × T
lnK = - 173.1 × 10³ J/mol / (8.314 J/mol.K) × 298 K
K = 4.542 × 10⁻³¹
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
Answer:
d. Heat is released from the reaction
Explanation:
A negative enthalpy change indicates that it is an exothermic reaction. Exothermic reactions release heat.
