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1 year ago

Which reaction displays an example of an Arrhenius base?

2 answers:

6 0

An Arrhenius base is a substance that, when dissolved in an aqueous solution, increases the concentration of hydroxide, or OH-, ions in the solution. I believe the correct answer from the choices listed above is option A. The reaction that displays an example of an Arrhenius base would be NaOH(s) --> Na+(aq) + OH–(aq). Hope this answers the question.

Katena32 [7]1 year ago

6 0

Answer:

answer is option A

Explanation:

just took the test and got it right

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ANSWER FOR: How did the lab activities help you answer the lesson question "How do the processes of conduction, convection, and

Alenkinab [10]

Answer: Conduction, convection, and radiation move energy from the Sun to Earth and throughout Earth.

Without more information about the experiment itself, I would choose the above answer as correct. All the other statements are correct, however none of them relates to the earth distribution processes on Earth. The last statement does.

8 0

1 year ago

Read 2 more answers

(a) What are the possible values of l for n = 4? (Enter your answers as a comma-separated list.)

vodka [1.7K]

Answer:

(a) 0,1,2,3 (b) -3,-2,-1,0,1,2,3 (c) 6 (d) 5

Explanation:

(a) for the principal quantum number 'n', the possible values of I = 0 to n-1. Thus, if the principal quantum number 'n' =4, I = 0,1,2,3.

(b) for a given number of 'I', the possible values of ml = -I to +I. Therefore, if I =3, then ml = -3,-2,-1,0,1,2,3

(c) 'I' which is the orbital angular momentum quantum number usually has values from 0,1,2,⋯,n−1. Therefore, for n greater than or 6, t would be greater than or equal to 5. Thus, the smallest possible value of n for which I can be 6 is 6.

(d) In a 3-dimensional figure, If the z-component of the orbital angular momentum Lz for which I=5 is measured, The possible outcomes will be:

mħ = -5ħ, -4ħ, -3ħ, -2ħ, -1ħ, 0, 1ħ, 2ħ, 3ħ, 4ħ, 5ħ.

Thus, the smallest possible l that can have a z component of 5ℏ is 5.

4 0

2 years ago

How many molecules of carbon dioxide are in 243.6 g of carbon dioxide?

german

Hey there !

Molar mass carbon dioxide:

CO2 = 44.01 g/mol

1) number of moles :

1 mole CO2 ------------- 44.01 g
(moles CO2) ------------ 243.6 g

moles CO2 = 243.6 * 1 / 44.01

moles CO2 = 243.6 / 44.01

=> 5.535 moles of CO2

Therefore:

1 mole -------------------- 6.02x10²³ molecules
5.535 moles ------------ ( molecules CO2)

molecules CO2 = 5.535 * ( 6.02x10²³) / 1

=> 3.33x10²⁴ molecules of CO2

3 0

1 year ago

Write a balanced half-reaction for the oxidation of liquid water H2O to aqueous hydrogen peroxide H2O2 in basic aqueous solution

nignag [31]

Answer : The balanced half-reaction in a basic solution will be,

$2OH^-(aq)\rightarrow H_2O_2(aq)+2e^-$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(OH^-)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The half reaction is :

$H_2O(l)\rightarrow H_2O_2(aq)$

Now balance the oxygen atoms.

$H_2O(l)\rightarrow H_2O_2(aq)+H_2O(l)$

Now balance the hydrogen atoms.

$H_2O(l)+2OH^-(aq)\rightarrow H_2O_2(aq)+H_2O(l)$

Now balance the charge.

$H_2O(l)+2OH^-(aq)\rightarrow H_2O_2(aq)+H_2O(l)+2e^-$

The balanced half-reaction in a basic solution will be,

$2OH^-(aq)\rightarrow H_2O_2(aq)+2e^-$

7 0

2 years ago

Según la Organización Mundial de la Salud, el nitrato de plata ( densidad = 4.35 g/cc ) es una sustancia con propiedades cáustic

OlgaM077 [116]

Answer:

737.52 mL de agua

Explanation:

En este caso solo debes usar la expresión de molaridad de una solución la cual es:

M = moles / V

Donde:

V: Volumen de solución.

Como queremos saber la cantidad de agua, queremos saber en otras palabras cual es la cantidad de solvente que se utilizó para preparar los 800 mL de disolución.

Una disolución se prepara con un soluto y solvente. El soluto lo tenemos, que es el nitrato de plata. Con la expresión de arriba, calculamos los moles de soluto, y luego su masa. Posteriormente, calculamos el volumen con la densidad, y finalmente podremos calcular el solvente de esta forma:

V ste = Vsol - Vsto

Primero calcularemos los moles de soluto:

moles = M * V

moles = 2 * 0.800 = 1.6 moles

Con estos moles, se calcula la masa usando el peso molecular reportado que es 169.87 g/mol:

m = moles * PM

m = 1.6 * 169.87 = 271.792 g

Ahora usando el valor de la densidad, calcularemos el volumen de soluto empleado:

d = m/V

V = m/d

V = 271.792 / 4.35

V = 62.48 mL

Finalmente, la cantidad de agua necesaria es:

V agua = 800 - 62.48

V agua = 737.52 mL

8 0

2 years ago