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enyata [817]
3 years ago
6

One way to represent a model of an atom is in a drawing. What is another way to show an atomic model? ​

Chemistry
1 answer:
mamaluj [8]3 years ago
7 0

I think by writing electronic configuration.

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In which of the titrations described below will the first (or only) equivalence point be reached upon the addition of 25.0 mL of
Mnenie [13.5K]

Answer:

1, 3, and 4

Step-by-step explanation:

We must calculate the volume of NaOH needed for each titration.

<em>1) HCl </em>

HCl + NaOH ⟶ NaCl + H₂O

n(HCl)       = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH)  = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

<em>2) H₂C₂O₄ </em>

H₂C₂O₄ + NaOH ⟶ NaHC₂O₄

n(H₂C₂O₄) = 50.0 mL × (0.100 mmol/1mL) = 5.00 mmol

n(NaOH)    = 5.00 mmol H₂C₂O₄ × (1 mmol NaOH/1 mmol H₂C₂O₄)

= 5.00 mmol NaOH

V(NaOH)   = 5.00 mmol × (1 mL/0.100 mmol) = 50.0 mL

<em>3) HC₂H₃O₂ </em>

HC₂H₃O₂ + NaOH ⟶ NaC₂H₃O₂ + H₂O

n(HC₂H₃O₂) = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

<em>4) HBr </em>

HBr + NaOH ⟶ NaBr + H₂O

n(HBr) = 25.0 mL × (0.100 mmol/1mL) =2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

Titrations 1, 3, and 4 reach the first or only equivalence point  at 25.0 mL NaOH.

8 0
3 years ago
Read 2 more answers
What is the total number of atoms in 1.0 mole of
emmainna [20.7K]

D. 18 x 10^23 is the total number of atoms in 1.0 mole of CO2
6 0
3 years ago
What happens to the individual molecules of a liquid as they gain enough kinetic energy to escape the surface of the liquid?
Bumek [7]
Im pretty sure They freeze
7 0
3 years ago
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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>
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3 years ago
Write and label the density triangle
Slav-nsk [51]

Answer:

It is basically a way of telling you how to solve for different variables in the equation d=m/v

Explanation:

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3 years ago
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