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nexus9112 [7]
3 years ago
9

Which of the following is true in regard to ions? A. A cation will hold a positive charge if it gains one or more electrons. B.

An ionized atom has a number of protons that is unequal to the number of electrons. C. Ions can carry only positive charges. D. Losing one or more electrons will turn an atom into an anion.
Chemistry
1 answer:
tigry1 [53]3 years ago
7 0

Answer:

B. An ionized atom has a number of protons that is unequal to the number of electrons.

Explanation:

For a neutral atom , the number of proton and electron is equal. An ionized atom has either loss or gain electron, thereby making the number of proton and electron unequal. The answer B is true because an ionized atom has either loss or gain electron to make the number of electron and proton unequal.

Option A is incorrect because a cations holds a positive charge when it loss one or more electron not when it gains one or more electron(s). Anions possess negative charge for gaining electron(s).

Option C is not true because ions can also carry negative charges and they are called anions.

Option D is false because losing one or more electron will turn an atom to a cations.

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It Is Called The Parent Nuclide

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Where would a primary producer be located in a food chain?
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Answer:

The primary producer would be at the bottom of the food chain.

Explanation:

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4. Cuál es la aplicación de las reacciones de óxido-reducción en la vida diaria y en la industria?
12345 [234]

Answer:

<em>Dentro de las aplicaciones de la óxido-reducción se pueden encontrar:</em>

  1. <u><em>La obtención del aluminio a partir de la alúmina y la electrolisis.</em></u>
  2. <u><em>La obtención de cloro, hidrógeno e hidróxido de sodio a partir del cloruro de sodio y la electrolisis.</em></u>
  3. <u><em>La combustión interna de un motor a gasolina u otro combustible fósil.</em></u>
  4. <u><em>Las termoeléctricas, las cuales para generar energía realizan combustión de carbón.</em></u>
  5. <u><em>La galvanoplastia, donde para evitar la corrosión de un metal se recubre con otro metal más resistente, por ejemplo: el recubrimiento del acero con zinc.</em></u>
  6. <u><em>La pilas o baterías de las cuales se obtiene energía química</em></u><em>.</em>

Explanation:

<em>Como puedes ver en la respuesta, la óxido-reducción tiene diversas aplicaciones en la vida moderna, desde todos los tipos de combustión los cuales sirven para brindar energía o movilizarte, hasta todas las funciones que se le ha dado a la electrolisis y a la obtención de la energía por medios químicos, incluso se puede considerar una aplicación de la óxido-reducción la incorporación de antioxidantes en los alimentos, los cuales disminuyen la velocidad de descomposición de los mismos. </em>

3 0
3 years ago
Calculate the molarity of a solution of sodium hydroxide, naoh, if 23.64 ml of this solution is needed to neutralize 0.5632 g of
GuDViN [60]
The  molarity  of  NaOH  needed  is  calculated   as  follows
calculate  the  moles  of  KhC8h4O4

that  is  moles  =  mass/molar  mass  of  KhC8h4O4(204.22 g/mol)

=0.5632g /204.22g/mol=  2.76  x10^-3  moles

write the  equation  for  reaction

khc8h4O4  +  NaOH  ---> KNaC8h4O4  +  H2O

from  the  equation  above  the   reacting  ratio   of   KhC8h4O4  to  NaOh  is  1:1  therefore  the  moles  of  Naoh  is  also  2.76  x10^-3  moles

molarity  of NaOh  =  (moles  of  NaOh /  volume ) x  1000

that  is { (2.76  x10^-3) / 23.64}  x100  =0.117 M
8 0
3 years ago
2. A 2.5 mol SAMPLE OF OXYGEN GAS (O2) INCREASES TO 3.2 mol
lana [24]

696.32 mmHg is the final pressure of the gas.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas.

Given data:

P_1 = 720 mmHg              

P_2 = ?

n_1 = 2.5 mol                

n_2 = 3.2 mol

V_1 = 34 L            

V_2 = 45 L

Formula

Combined gas law

\frac{P_1 V_1}{n_1}  = \frac{P_2 V_2}{n_2}

P_2 = 696.32 mmHg

Hence, 696.32 mmHg is the final pressure of the gas.

Learn more about an ideal gas equation here:

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