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3 years ago

Determine the rate of a reaction that follows the rate law rate = k a m b i where

1 answer:

6 0

(missing part of your question):
when we have K = 1 x 10^-2 and [A] = 2 M & [B] = 3M & m= 2 & i = 1
So when the rate = K[A]^m [B]^i
and when we have m + i = 3 so the order of this reaction is 3 So the unit of K is L^2.mol^-2S^-1
So by substitution:
∴ the rate = (1x 10 ^-2 L^-2.mol^-2S^-1)*(2 mol.L^-1)^2*(3mol.L^-1)
= 0.12 mol.L^-1.S^-1

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How does an electron reach an excited state

quester [9]

When an electron quickly occupies an strength state increased than its ground state, it is in an excited state. An electron can end up excited if it is given greater energy, such as if it absorbs a photon, or packet of light, or collides with a close by atom or particle.

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2 years ago

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A student ran the following reaction in the laboratory at 671 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 7.33×10-2 moles of N

vaieri [72.5K]

Answer:

Kc = 8.05x10⁻³

Explanation:

This is the equilibrium:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

Initially 0.0733

React 0.0733α α/2 3/2α

Eq 0.0733 - 0.0733α α/2 0.103

We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.

Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.

3/2α = 0.103

α = 0.103 . 2/3 ⇒ 0.0686

So, concentration in equilibrium are

NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682

N₂ = 0.0686/2 = 0.0343

So this moles, are in a volume of 1L, so they are molar concentrations.

Let's make Kc expression:

Kc= [N₂] . [H₂]³ / [NH₃]²

Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³

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2 years ago

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2 years ago

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calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.

Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

$P_w=P_p+P_s$

$P_w=P_p+(-iCRT)$

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= $P_w$

$P_p=0 bar$ (At standard temperature)

$P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar$

$P_w=0 bar+(-3.406 ) bar$

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

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2 years ago

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Hello!
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2 years ago