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GaryK [48]
3 years ago
7

Un hidrocarburo tiene como composición en masa: C= 85.8% ; H= 14.2% . Como dato nos brindan que una molécula de este compuesto p

esa: LaTeX: 2.1x10^{-22}g2.1 x 10 − 22 g. Hallar la fórmula molecular. Datos: P.A (uma): C=12, H=1
Chemistry
1 answer:
OLga [1]3 years ago
8 0

Answer:

C_{9.042}H_{17.958}

Explanation:

(This exercise has been written in Spanish and for that reason explanation will be held in such language)

Por la definición del Número de Avogadro, un mol del compuesto contiene 6.022 \times 10^{23} moléculas. La masa de un mol se determina mediante regla de tres simple:

x = \frac{6.022\times 10^{23}\,molecules}{1\,molecule} \times (2.1\times 10^{-22}\,g)

x = 126.462\,g

Las participaciones másicas del carbono y el hidrógeno son, respectivamente:

m_{C} = 0.858\cdot (126.462\,g)

m_{C} = 108.504\,g

m_{H} = 126.462\,g - 108.504\,g

m_{H} = 17.958\,g

Ahora, el número de moles de cada elemento es calculado a continuación:

n_{C} = \frac{108.504\,g}{12\,\frac{g}{mol} }

n_{C} = 9.042\,mol

n_{H} = \frac{17.958\,g}{1\,\frac{g}{mol} }

n_{H} = 17.958\,mol

La fórmula molecular del hidrocarburo es:

C_{9.042}H_{17.958}

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For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:
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Answer:

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c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

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At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

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T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

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Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

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For an endothermic reaction, the products will be closer in energy to the transition state than what the reactans will be; so, the activation energy of the reversed reaction is lower than the activation energy of the forward reaction.

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Activation energy of reverse rxn = Activation energy of forward rxn - ΔH rxn

=> Activiation energy of reverse rxn = 102 kJ/mol - 55 kJ/mol = 47 kJ/mol

Answer: 47 kJ/mol
8 0
3 years ago
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