The total number of how many objects you have

Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286

ra1l [238]

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.

Now the balanced equation is:

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.

Now based on the balanced chemical reaction,

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.

Thus, the maximum work, which can obtained is 1332 kJ.

6 0

3 years ago

The pH of a 0.15 M butylamine, C&H3NH2 solution is 12.0 at 25°C. Calculate the dissociation

Dmitrij [34]

The dissociation constant of the base : 7.4 x 10⁻⁴

<h3>Further explanation</h3>

Butylamine, C4H9NH2 Is A Weak Base

Kb is the dissociation constant of the base.

LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)

$\rm Kb=\dfrac{[L][OH^-]}{[LOH]}$

[OH⁻] for weak base can be formulated :

$\tt [OH^-]=\sqrt{Kb.M}$

pH of solution : 12

pH+pOH=14, so pOH :

14-12 = 2, then :

$\tt [OH^-]=10^{-pOH}\\\\(OH^-]=10^{-2}$

the the dissociation constant (Kb) =

$\tt 10^{-2}=\sqrt{Kb.0.15}\\\\10^{-4}=Kb\times 0.15\\\\Kb=\dfrac{10^{-4}}{0.15}=6.6\times 10^{-4}$

Or you can use from ICE method :

C4H9NH2(aq) + H2O(l) ⇌ C4H9NH3+(aq) + OH-(aq)

0.15

x x x

0.15-x x x

$\tt Kb=\dfrac{x^2}{0.15-x}\rightarrow x=[OH^-]\\\\Kb=\dfrac{10^{-4}}{0.15-10^{-2}}=7.14\times 10^{-4}$

6 0

3 years ago

Please please please please please help me

Papessa [141]

Answer: Iodine

Explanation:

6 0

4 years ago

Which question must be answered to complete the table below?

I am Lyosha [343]

I need more details so I can answer it for you

8 0

4 years ago

How many atoms (mol) is in 10g of sodium, sodium chloride and Iron chloride?

SashulF [63]

Answer:

See explanation

Explanation:

Molar mass of NaCl = 58.5 g

Number of moles contained in 10 g of NaCl = 10 g/58.5 g = 0.17 moles

If 1 mole of NaCl contains 6.02 * 10^23 atoms

0.17 moles of NaCl contains 0.17 * 6.02 * 10^23 atoms = 1.02 * 10^23 atoms

Molar mass of Fe II chloride = 126.751 g/mol

Number of moles = 10 g/126.751 g/mol = 0.0789 moles

Number of atoms = 0.0789 moles * 6.02 * 10^23 atoms = 4.7 * 10^22 atoms

Molar mass of Na = 23 g/mol

Number of moles = 10g/23 g/mol = 0.43 moles

Number of atoms = 0.43 moles * 6.02 * 10^23 atoms = 2.59 * 10^ 23 atoms

8 0

3 years ago