Explain why a mixture is used as the electrolyte??

Consider the reaction. 2c(s,graphite) h2= c2h2(g) Based on the equation and the information in the table, what is the enthalpy o

xxTIMURxx [149]

<h3>The enthalpy of the reaction : C.226.73 kJ</h3><h3>Further explanation</h3>

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

Delta H reaction (ΔH) is the difference between product enthalpy and reactant enthalpy (Hp-Hr)

Reaction :

2C + H2 -> C2H2

(ΔH) reaction = (ΔH) product - (ΔH) reactants

(ΔH) reaction = (ΔH) C2H2 - ((ΔH) C + (ΔH) H2)

for free element (ΔH) it has zero value, so (ΔH) C and (ΔH) H2 have zero value

From the available data: Hf (kJ / mol) C2 H2 (g)= 226.73

then

(ΔH) reaction = ΔH) C2H2 - ((ΔH) C + (ΔH) H2)

(ΔH) reaction = 226.73 - (0 + 0)

(ΔH) reaction = 226.73 (answer C)

<h3>Learn more</h3>

(ΔH) reaction

brainly.com/question/1889660

Keywords : (ΔH) reaction, C2H2

8 0

3 years ago

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A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,

Ad libitum [116K]

The answer is 3.68 L

8 0

3 years ago

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A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2

sukhopar [10]

Answer:

a) $\triangle G^{0} = 7.31 kJ/mol$

b) $K_{-1} = 0.0594 m^{-1} s^{-1}$

Explanation:

Equation of reaction:

$2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)$

Initial pressure 3 1 0

Pressure change 2P 1P 2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

$K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }$

$K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}$

Standard free energy:

$\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol$

b) value of k−1 at 27 °C, i.e. 300K

$K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}$

$K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}$

$K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}$

6 0

3 years ago

Definition of compound inequality.

Hitman42 [59]

A compound inequality<span> is an equation with two or more </span>inequalities<span> joined together with either</span>

6 0

3 years ago

A geological sample is found to have a Pb-206/U-238 mass ratio of 0.337/1.00. Assuming there was no Pb-206 present when the samp

azamat

Answer:

$Age=2.52*10^9 years$

Explanation:

To determine the age, we need to know how much U have become Pb, so first we have to calculate the Pb moles present in a sample:

$n_{Pb}=\frac{0.337g}{206g/mol}=0.00164 mol$

$n_{U}=\frac{1 g}{238g/mol}=0.0042 mol$

The percentage of U degradation:

$P=\frac{n_{Pb}}{n_{Pb}+n_{U}}$

$P=\frac{0.00164}{0.00164+0.0042}=0.28$

Assuming that the life time is linear:

$Age=\frac{4.5*10^9 years}{0.5 life time}*0.28 life time$

$Age=2.52*10^9 years$

4 0

3 years ago

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