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Galina-37 [17]
3 years ago
13

According to the collision theory, which is required for a high number of effective collisions? a very low amount of force from

colliding molecules a very low amount of kinetic energy from colliding molecules molecular collisions that have very specific orientations molecular collisions with energy to overcome activation energy
Chemistry
2 answers:
ra1l [238]3 years ago
5 0

Answer:

the answer is D

Explanation:

i answered c and got it wrong on edg... i hope this helps :)

neonofarm [45]3 years ago
4 0

Answer:

molecular collisions that have very specific orientations

Explanation:

  • Based on the collision theory, a high frequency of effective collision is dependent on the molecular collisions that have very specific orientations.
  • The collision theory suggests that for reactions to occur, there must collision between reacting particles.
  • The number of collision is dependent on the number of collision per unit time as well as fractions of effective collision.
  • To attain effective collision, colliding particles must be properly oriented to give the desired product.

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Determine the value of the equilibrium constant, Kgoal, for the reaction
hodyreva [135]

Answer:

Part A

K_{goal}= 3.22\times 10 ^{-66}

Part B

K_{goal}= 2.26\times10^{-21}

Explanation:

Part A:

The given equation is not balanced. So, initially let's balance the equation by taking 24 moles of each of the reagent and NO.

24N₂ + 24H₂O ⇌ 24NO + 12N₂H₄

Now, simplify the equation by dividing both sides by 12. The final balanced equation is the following

2N₂ + 2H₂O ⇌ 2NO + N₂H₄

The above-balanced equation can be solved algebraically to obtain the required Kgoal value.

Adding given equations 1, 2 and 3 we obtain the required equation.

When the equations are added, the equilibrium constants of each equation are multiplied. Mathematically it can be represented as,

K_1 \times K_2 \times K_3 = K_{goal}

K_{goal}= 4.10 \times 10^{-31} \times 7.40 \times 10^{-26} \times 1.06 \times 10^{-10}

K_{goal}= 3.22\times 10 ^{-66}

Part B:

The required equation is balanced, Now

Let.

P₄(s)+6Cl₂(g) ⇌ 4PCl₃(g), K₁=2.00×10¹⁹ ------------------------------------ (a)

PCl₅(g) ⇌ PCl₃(g)+Cl₂(g), K₂=1.13×10⁻² --------------------------------------- (b)

By multiplying equation 2 by 4 and subtracting equation 1 from it, we get

4PCl₅(g) ⇌ P₄(s)+10Cl₂(g)

The Kgoal for the above equation is the product of four times K₂ and inverse K₁ according to the applied operation. Mathematically,

K_{goal}= 4K_2 \times \frac{1}{K_1}

K_{goal}= 4(1.13\times10^{-2}) \times \frac{1}{2.00\times10^{19}}

K_{goal}= 2.26\times10^{-21}

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3 years ago
Water has a higher specific heat index than other common materials, so it takes a lot of energy to raise the temperature of wate
Maurinko [17]

Answer: Areas surrounded by water would have less of a change in temperature

throughout the year.

Explanation: Water has a higher specific heat index so it's temperature will be slow to rise. Water also holds heat longer than most material hence areas sorrounsing water will have less temperature change throughout the year.

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4 years ago
Banana dipped in liquid experiments
aleksley [76]
Https://www.youtube.com/watch?v=LXDG0nqYdR4
4 0
4 years ago
Calculate E o , E, and ΔG for the following cell reactions (a) Mg(s) + Sn2+(aq) ⇌ Mg2+(aq) + Sn(s) where [Mg2+] = 0.025 M and [S
Rudik [331]

E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

<u>Explanation:</u>

Mg(s) + Sn²⁺(aq) ⇌ Mg²⁺(aq) + Sn(s)

[Mg2+] = 0.025 M

[Sn2+] = 0.040 M

First we need the standard reduction potentials:

. . . . . . . . . . . . . . . . . E°(V)

Mg²⁺ + 2 e⁻ ⇌ Mg(s). . .−2.372

Sn²⁺ + 2 e⁻ ⇌ Sn(s) . . . −0.13

Take the more negative (or less positive in other cases) one, and write it as an oxidation:

Mg(s) ⇌ Mg²⁺ + 2 e⁻. . .+2.372 V

Combine them,

Mg(s) + Sn²⁺ ⇌ Mg²⁺ + Sn(s)

E°(cell) = +2.372 – 0.13 V = 2.24 V

To get the cell potential under the conditions given, use the Nernst Equation:

E(cell) = E°(cell) – [(0.059)/n]•logQ = 2.24 V – 0.0295 V • log [Mg²⁺]/[Sn²⁺]

Note that the solids don't appear in Q, only the concs. of the dissolved ions.

E(cell) = 2.24 V – 0.0295 V X log (0.025)/(0.040)

          = 2.24 + 0.006 V ≈ 2.246 V

The concentration ratio in Q (Sn²⁺ and Mg²⁺) is too close to 1 to shift E(cell) significantly from E°(cell) given the precision I have for the Sn reduction potential.

∆G = –nFE(cell) = –2(96.485 kJ/mol•V)(2.246 V) = –433 kJ/mol

E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

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BartSMP [9]
This would be C, 1.044 mi/min
5 0
3 years ago
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