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Novosadov [1.4K]

3 years ago

14

Based on the activity series provided, which reactants will form products? F > Cl > Br > I CuI2 + Br2 Right arrow. Cl2

+ AlF3 Right arrow. Br2 + NaCl Right arrow. CuF2 + I2 Right arrow.

2 answers:

Sonja [21]3 years ago

8 0

Answer:

i believe the answer is a

Explanation:

agasfer [191]3 years ago

7 0

Answer: Cul2 + Br2 ->

Explanation:

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Explanation:

The given data is as follows.

$\Delta H$ = 286 kJ = $286 kJ \times \frac{1000 J}{1 kJ}$

= 286000 J

$S_{H_{2}O} = 70 J/^{o}K$ , $S_{H_{2}} = 131 J/^{o}K$

$S_{O_{2}} = 205 J/^{o}K$

Hence, formula to calculate entropy change of the reaction is as follows.

$\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)$

= $[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]$

= $[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]$

= 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

$\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}$

= $286000 J - (163.5 J/K \times 298 K)$

= 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

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Read 2 more answers

At the start of a reaction, there are 0.0249 mol N2,

gladu [14]

Answer:

Explanation:

The reaction is given as:

$N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$

The reaction quotient is:

$Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}$

From the given information:

TO find each entity in the reaction quotient, we have:

$[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}$

$[N_2] = \dfrac{0.024 }{3.5}$

$[N_2] = 0.006857$

$[H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}$

$[H_2] = 9.17 \times 10^{-3}$

∴

$Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135$

However; given that:

$K_c = 1.2$

By relating $Q_c \ \ and \ \ K_c$ , we will realize that $Q_c \ \ < \ \ K_c$

The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.

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