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OleMash [197]
2 years ago
13

What is the percent composition of nitrogen in N 2 O

Chemistry
1 answer:
mash [69]2 years ago
5 0

Answer:

63.6%

Explanation:

The given compound is:

     N₂O;

The problem here is to find the percent composition of nitrogen in the compound.

First find the molar mass of the compound:

 Molar mass of N₂O = 2(14) + 16  = 44g/mol

So;

 Percentage composition of Nitrogen  = \frac{2 x 14}{44}  x 100  = 63.6%

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Answer:

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Explanation:

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A first order reaction has a rate constant of 0. 543 at 25°C. Given that the activation energy is 75. 9 kj/mol. Calculate the ra
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The rate constant of first order reaction at 32. 3 °C is 0.343 /s must be less the 0. 543 at 25°C.

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The value of ok suggests the equilibrium ratio of products to reactants. In an equilibrium combination both reactants and merchandise co-exist. big ok > 1 merchandise are k = 1 neither reactants nor products are desired.

Rate constant K₁  = 0. 543 /s

T₁  = 25°C

Activation energy Eₐ =  75. 9 k j/mol.

T₂ = 32. 3 °C.

K₂ =?

formula;

log K₂/K₁=  Eₐ /2.303 R [1/T₁ - 1/T₂]

putting the value in the equation  

K₂ = 0.343 /s

Hence, The rate constant of first order reaction at 32. 3 °C is 0.343 /s

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1 year ago
Un móvil se mueve con movimiento acelerado. En los segundo 2 y 3 los espacios recorridos son 90 y 120 m, Calcula la velocidad in
faust18 [17]

Answer:

La velocidad inicial es 55 \frac{m}{s}y su aceleración es -10 \frac{m}{s^{2} }

Explanation:

Un movimiento es rectilíneo uniformemente variado, cuando la trayectoria del móvil es una línea recta y su velocidad  varia la misma cantidad en cada unidad de tiempo . Dicho de otra manera, este movimiento se caracteriza por una trayectoria que es una línea recta y la velocidad cambia su módulo de manera uniforme: aumenta o disminuye en la misma cantidad por cada unidad de tiempo. Y la aceleración es constante y no nula (diferente de cero).

En este caso la posición del objeto esta dada por la expresión:

x=x0+v0*t+\frac{1}{2} *a*t^{2}

donde x es la posición del cuerpo en un instante dado, x0 la posición en el instante inicial, v0 la velocidad inicial y a la aceleración.

En este caso, por un lado podes considerar:

  • x= 90 m
  • x0= 0 m
  • v0= ?
  • t= 2
  • a= ?

Reemplazando obtenes:

90=v0*2+\frac{1}{2} *a*2^{2}

90=v0*2+\frac{1}{2} *a*4

90=v0*2+2*a

Y por otro lado tenes:

  • x= 120 m
  • x0= 0
  • v0= ?
  • t= 3
  • a= ?

Reemplazando obtenes:

120=v0*3+\frac{1}{2} *a*3^{2}

120=v0*3+\frac{1}{2} *a*9

120=v0*3+\frac{9}{2} *a

Por lo que tenes el siguiente sistema de ecuaciones:

\left \{ {{2*v0+2*a=90} \atop {3*v0+\frac{9}{2} *a=120}} \right.

Resolviendo por el método de sustitución, que consiste en aislar en una ecuación una de las dos incógnitas para sustituirla en la otra ecuación, obtenes:

Despejando v0 de la primera ecuación:

v0= \frac{90-2*a}{2}

Reemplazando en la segunda ecuación:

120=\frac{90-2*a}{2} *3+\frac{9}{2} *a

Resolviendo:

120=(90-2*a)*\frac{3}{2} +\frac{9}{2} *a

120=135-3*a +\frac{9}{2} *a

120-135=-3*a +\frac{9}{2} *a

-15=\frac{3}{2} *a

\frac{-15}{\frac{3}{2} } =a

-10=a

Reemplazando el valor de a en la expresión despejada anteriormente obtenes:

v0= \frac{90-2*(-10)}{2}

Resolviendo:

v0= \frac{90+20}{2}

v0= \frac{110}{2}

v0=55

<u><em>La velocidad inicial es 55 </em></u>\frac{m}{s}<u><em>y su aceleración es -10 </em></u>\frac{m}{s^{2} }<u><em></em></u>

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