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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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Explanation:

4 0
3 years ago
Atmospheric pressure yesterday was given as 0.875 atm. Express this pressure in units of torr and also in kilopascals (kPa), res
Aneli [31]
This problem is about conversion and dimensional analysis. Important information to know:
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For atm to torr conversion:

0.875 atm * (760 torr / 1 atm) = 665 torr

For atm to kPa conversion:

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5 0
3 years ago
What is the molar mass of an unknown gas with a density of 2.00 g/L at 1.00 atm and 25.0 °C?
soldier1979 [14.2K]

Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

P

V

=

n

R

T

a

a

∣

∣

−−−−−−−−−−−−−−−

, where

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V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

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Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

ρ

, under those conditions for pressure and temperature, and its molar mass,

M

M

.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

m

of this gas, you can express its molar mass as the ratio between

m

and

n

, the number of moles it contains

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

M

M

=

m

n

a

a

∣

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

ρ

=

m

V

a

a

∣

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

M

⋅

R

T

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P

V

⋅

M

M

=

m

⋅

R

T

P

⋅

M

M

=

m

V

⋅

R

T

M

M

=

m

V

⋅

R

T

P

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(

2

)

to write

M

M

=

ρ

⋅

R

T

P

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M

M

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1.02

g

L

⋅

0.0821

atm

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L

mol

⋅

K

⋅

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273.15

+

37

)

K

0.990

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∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

26.3 g mol

−

1

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