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Aliun [14]
3 years ago
12

Formula for the elements if they were to bong together. Iodine and Bromine

Chemistry
2 answers:
blondinia [14]3 years ago
7 0

Answer:

I(Br)

Explanation:

You will be combining one iodine atom with one bromine atom.

Usimov [2.4K]3 years ago
6 0
Lo voy a escribir en español , perdón no se
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The more the energy, the larger the (a,b,c,d)?
Elanso [62]

Answer: D:wavelenght

Explanation: Students will understand that shorter wavelengths have higher frequency and energy.

3 0
3 years ago
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Describe how particles of iron (Fe) an oxygen (O2) react to how produce oxide (Fe2O3), also known as rust.
forsale [732]

Answer:

Described by a redox reaction below

Explanation:

Iron(III) oxide is an ionic compound, since it consists of a metal, iron, and a nonmetal, oxygen.

Ionic compounds are formed when metals lose their valence electrons in order to have an octet in their previous shell and donate them to nonmetal atoms, so that nonmetals fill their outer shell to have an octet.

As a result, positive ions (cations) and negative ions (anions) are formed. When iron reacts with oxygen, the following reaction takes place:

4 Fe (s) + 3 O_2 (g)\rightarrow 2 Fe_2O_3 (s)

This is a redox (oxidation–reduction) reaction, since we have electron loss and gain. Four iron atoms lose a total of 12 electrons to obtain a +3 charge in the final compound, while 3 oxygen molecules gain these 12 electrons to become 6 oxide anions with a -2 charge.

4 0
3 years ago
Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are
sertanlavr [38]

Answer:

Reaction A and B are unfavorable.

Explanation:

Gibbs free energy is an energy which that can be use to convert into useful work.

ΔG°=ΔH°-TΔS°

ΔG°= Gibbs free energy

ΔH° =  enthalpy of reaction

T = temperature of eh reaction

ΔS° = Entropy change

  • If the Gibbs  free energy of the reaction is positive than the reaction will be non spontaneous and the chemical reaction will be not feasible.
  • If the Gibbs  free energy of the reaction is negative than the reaction will be spontaneous and the chemical reaction will be feasible .

According to given information in the question:

Reaction A and B are non spontaneous as their Gibbs free energy value is positive.hence both are unfavorable.

3 0
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>

3 0
3 years ago
What scientist described a positively charged core in the middle of the atom
hjlf
Chadwick, Thompson, Rutherford, Bohr
8 0
4 years ago
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