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zloy xaker [14]
3 years ago
14

En relación a la escala de ph se puede decir lo siguiente: I. Mientras más cerca del 8 la sustancia es más básica II. El papel t

ornasol es un papel indicador, que al generar color azul predice que la sustancia es ácida. III. El agua marina es un ejemplo de sustancia básica
Chemistry
1 answer:
yawa3891 [41]3 years ago
3 0

Answer:

Las sustancias son mas basicas a medida que su ph se acerca a 7 y lo supera.(esto quiere decir que la afirmacion de la pregunta no es la correcta)

En cuanto al papel tornasol, tambien es falsa esa afirmacion ya que este cambia de color rojo-fuxia a azul cuando la sustancia quimica pasa de acida a basica, por ende cuando se alcanza ph de 7 para arriba.

El agua marina es un ejemplo de sustancia ALCALINA, su gran contenido sal-mineral hace que sea sumamente alcalina, superando altamente el ph7, es decir lo contrario al jugo de limon por ejemplo que es acido y ronda en un ph 3,5.

Explanation:

Las sustancias acidas son aquellas que tienen un valor de ph de 0 a 7, a medida que estas acerquen su valor de ph a 7 menos acidez tienen.

En cuanto a las basicas o neutras sus ph rondan en los 7, una vez superado ampliamente este valor como por ejemplo ph10 se considerara que la solucion es alcalina.

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Bromine has an atomic mass of 79.9 amu, but only has 2 naturally occurring isotopes, Bromine-79 and Bromine-81. If Bromine-80 do
galina1969 [7]

Educated Guess Here!

Since Br-80 does not exist, maybe that means Br-79 or Br-81 have very unequal abundances. For example, Br-79 may have 75% abundance whereas Br-81 may have 25% abundance.

5 0
3 years ago
What is work and state the law of conservation mass?<br>​
Butoxors [25]

Answer:

Work is a force causing the movement or displacement of an object

law of conservation mass:

1. Atoms cannot be created or destroyed in a chemical reaction.

2. Molecules cannot be created or destroyed in a chemical reaction.

3. Compounds cannot be created or destroyed in a chemical reaction.

4. Heat cannot be created or destroyed in a chemical reaction.

7 0
3 years ago
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Please refer to image please
GarryVolchara [31]

Answer:

76.9L

Explanation:

Based on the graph, whenever the temperature increases by 100K, the volume increases by 10L, so do 769/10

7 0
2 years ago
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
3 years ago
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