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kicyunya [14]
3 years ago
6

How would you prepare 50.0 ml of 0.600 m hno3 from a stock solution of 4.00 m hno3? what volume (in ml) of the 4.00 m hno3 solut

ion will you need?
Chemistry
1 answer:
serg [7]3 years ago
5 0
It will need 40000ml of the solution.
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Two trays of lasagna are identical except for their thermal energies.
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Answer:

the one with less thermal energy

Explanation:

thermal energy is heat

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Highlight the basic points in Lewis and Langmuir theory of electrovalency
ankoles [38]
The correct answer for the question that is being presented above is this one: "Electrovalency is characterized with the transferring of one or more electrons from one atom to another together with the formation of ions and as well as the number of positive and negative charges. 

The Lewis and Langmuir theory of electrovalency (and as well as Kossel's) is dealing with Ionic bonds. 

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<span>The Lewis-Langmuir electron-pair or covalent bond is referred as the homopolar bond, where the complete transfer of electrons give rise to ionic, or electrovalent bond (1) through attraction of opposite charges.</span>
8 0
3 years ago
What is the molarity of a solution in which 58g of nacl are dissolved in 1.0 l of solution
Lady_Fox [76]

Hey there :

Molar mass of NaCl = 58.44 g/mol

Number of moles :

n = mass of solute / molar mass

n = 58 / 58.44

n = 0.9924 moles of NaCl

Volume = 1.0 L

Therefore:

Molarity = number of moles / volume ( L )

Molarity = 0.9924 / 1.0

Molarity = 0.9924 M

Hope that helps!

7 0
3 years ago
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
3 years ago
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