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iris [78.8K]
2 years ago
10

The formula equation of Acetylene + oxygen ----> carbon dioxide + water ​

Chemistry
1 answer:
snow_lady [41]2 years ago
8 0

Answer:

The final balanced equation is : 2C2H2+5O2→4CO2+2H2O.

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Determine el PH y el % de disociación de una solución de ácido débil, sabiendo que se disuelven 20 gramos del ácido (masa molar=
IceJOKER [234]

Answer:

pH = 4.27. Porcentaje de disociación: 0.03%

Explanation:

El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Donde la constante de equilibrio, Ka, es

Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]

Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:

[H⁺] = [X⁻]

[HX] es:

20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M

Reemplazando es Ka:

1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]

2.858x10⁻⁹ = [H⁺]²

5.35x10⁻⁵M = [H⁺]

pH = -log[H⁺]

<h3>pH = 4.27</h3>

El porcentaje de disociacion es [X⁻] / [HX] inicial * 100

Reemplazando

5.35x10⁻⁵M / 0.1732M * 100

<h3>0.03%</h3>
5 0
2 years ago
Choose two of the following scientists: Anton Lavoisier, John Dalton, JJ Thomson, Robert Millikan, Ernest Rutherford, James Chad
OverLord2011 [107]

Ernest Rutherford

J. J Thomson

Explanation:

<u>Ernest Rutherford</u>

In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.

 Experiment

 In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.

 Discovery and reflection on the atomic theory

To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.

<u>J. J Thomson</u>

Experiment

In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.

Discovery and reflection on the atomic theory

From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:

  • they move in a straight line
  • they possess kinetic energy
  • they attract positive charges and repels negative charges

Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.

learn more:

Rutherford brainly.com/question/1859083

#learnwithBrainly

6 0
3 years ago
Which of the following best describes an example of applied chemistry?
hram777 [196]

Answer:

Explanation:

Examples of applied chemistry include creation of the variety of laundry detergents on the market and development of oil refineries.

4 0
2 years ago
What is the concentration of the barium hydroxide solution if 50.0 mL of a 0.425 M HNO3 solution is required to neutralize a 36.
fomenos

The molarity of Barium Hydroxide is 0.289 M.

<u>Explanation:</u>

We have to write the balanced equation as,

Ba(OH)₂ + 2 HNO₃ → Ba(NO₃)₂ + 2 H₂O

We need 2 moles of nitric acid to react with a mole of Barium hydroxide, so we can write the law of volumetric analysis as,

V1M1 = 2 V2M2

Here V1 and M1 are the volume and molarity of nitric acid

V2 and M2 are the volume and molarity of Barium hydroxide.

So the molarity of Ba(OH)₂, can be found as,

$ M2 = \frac{V1 \times M1}{2 \times V2}

   $M2 = \frac{50 \times 0.425 }{2 \times 36.8}

      = 0.289 M

5 0
3 years ago
If the solubility of KCl in 100 mL of H₂O is 34 g at 20 °C and 43 g at 50 °C, label each of the following solutions as unsaturat
sertanlavr [38]

Answer:

a) Unsaturated

b) Supersaturated

c) Unsaturated

Explanation:

A saturated  solution contains the <u>maximum amount of a solute that will dissolve in a given  solvent at a specific temperature</u>.

An unsaturated solution contains <u>less solute than it  has the capacity to dissolve. </u>

A supersaturated solution, <u>contains more  solute than is present in a saturated solution</u>. Supersaturated solutions are not very  stable. In time, some of the solute will come out of a supersaturated solution as crystals.

According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:

a) 30 g in 100 mL of H₂O at 20 °C  ⇒ unsaturated

b) 65 g in 100 mL of H₂O at 50 °C  ⇒ supersaturated

c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)

8 0
3 years ago
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