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julsineya [31]
3 years ago
12

Which of the following is not an indication of a chemical reaction? a substance dissolves formation of a precipitate color chang

e the container gets cold
Physics
2 answers:
yKpoI14uk [10]3 years ago
6 0

Answer:

A substance dissolves

SVETLANKA909090 [29]3 years ago
4 0

a substance dissolves.  


like adding a soluble salt to water, it just dissolves, i.e dissociates homogeneously as water is able to dissociate salts (ionic compounds) into its ions. (it can also dissociate other non-ionic compounds like HCL)  


the salt still remains chemically as a salt and is unchanged chemically thus it is not an indication of a chemical reaction as no chemical reaction has taken place.  


the formation of a precipitate is a chemical reaction because a new substance (i.e new chemical) is formed. For example adding aqueous sodium hydroxide into an aqueous solution with CU2+ cations will form a blue precipitate (that is copper (II) hydroxide which is insoluble, hence it precipitates). Since a new chemical is formed, a chemical reaction has taken place and thus indicates a chemical reaction.  


color change... im not sure but usually a color change will only occur when a new substance is formed. Like iron corrodes (i.e rust) slowly in moist air to form hydrated iron (III) oxide that is rust. (brown color).  


usually adding a mixture to a mixture has little energy change, i.e little heat taken in by the reaction mixture or little heat given out by the reaction mixture. Whereas when a new substance is formed, there is usually noticeable energy change like the container gets colder or hotter (without heat being supplied of course). For example dissolving basic oxides into water releases energy ( more energy released than gained = exothermic reaction).  


i think that should be the answer... hope it helped :D

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Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree
kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    I_f_3 = 27.57 W/cm^2

2 When third  polarizer is removed the intensity after it passes through the stack is    

                I_f_2 = 102.24 W/cm^2

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  \theta_2 = 21^o  

        The angle of the third  polarizing to the first is     \theta_3 = 61^o

        The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

Let the initial intensity of the beam of light before polarization be I_p

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     I_1 = \frac{I_p}{2}

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    I_2 = I_1 cos^2 \theta_1

                       = \frac{I_p}{2} cos ^2 \theta_1

The intensity of light that will emerge from the third filter is mathematically represented as

                  I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                          I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

making I_p the subject of the formula

                  I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

    Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                           =234.622W/cm^2

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

I_f_3 is the intensity of the light emerging from the stack

                     

substituting values

                     I_f_3 = \frac{234.622}{2} * cos^2(61)

                       I_f_3 = 27.57 W/cm^2

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

I_f_2 is the intensity of the light emerging from the stack

Substituting values

                  I_f_2 =  \frac{234.622}{2} cos^2 (21)

                     I_f_2 = 102.24 W/cm^2

   

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kotegsom [21]

Answer : The specific heat of aluminum is, 0.90J/g^oC

Solution : Given,

Heat absorbs  = 677 J

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Initial temperature = 50^oC

Formula used :

Q= m\times c\times \Delta T

or,

Q= m\times c\times (T_{final}-T_{initial})

Q = heat absorbs

m = mass of the substance

c = heat capacity of aluminium

T_{final} = final temperature

 T_{initial} = initial temperature

Now put all the given values in the above formula, we get the specific heat of aluminium.

677g= (10g)\times c\times (125-50)^oC

c=0.9026J/g^oC=0.90J/g^oC

Therefore, the specific heat of aluminum is, 0.90J/g^oC

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Answer:

40 metros en 10 segundos.

Explicación:

Una grúa está levantando una caja de 1000 kg atada a una cadena y la caja, que inicialmente está en reposo, aumenta su velocidad en 4 m / s por segundo, por lo que si la región de colocación está a 40 metros de distancia, la grúa tarda 10 segundos en completar el proceso. tarea de movimiento y colocación de la caja. La grúa se utiliza para levantar y mover cargas pesadas, máquinas, materiales y mercancías para diferentes propósitos. Entonces el trabajo realizado por la grúa depende de la velocidad y la distancia de colocación.

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3 years ago
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