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Radda [10]
2 years ago
10

The elemental composition of a compound is determined to be 39.5% oxygen, 32.1% chromium and 28.4% sodium by weight.

Chemistry
1 answer:
maksim [4K]2 years ago
3 0

Answer:

Explanation:

To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

​%C=

​12.04g compound

​

​7.34g C

​​ ×100%=61.0%

​%H=

​12.04g compound

​

​1.85g H

​​ ×100%=15.4%

​%N=

​12.04g compound

​

​2.85g N

​​ ×100%=23.7%

​​  

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

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Sodium react gently with water ture or false<br>​
Maksim231197 [3]

false,

sodium has a vigorous reaction with water.

This can be observed by putting a piece of sodium in water,

it moves swiftly across the surface of water and reacts violently.

5 0
3 years ago
Explain hydrogen bonding.
denis23 [38]

Answer:

Hydrogen bonding, interaction involving a hydrogen atom located between a pair of other atoms having a high affinity for electrons; such a bond is weaker than an ionic bond or covalent bond but stronger than van der Waals forces. Hydrogen bonds can exist between atoms in different molecules or in parts of the same molecule.

Explanation:

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Identify the 2 numbers given for each element
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where are the element you haven't showed it so that we can answer it

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3 years 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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Yes explosion is mainly energy 

5 0
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