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Travka [436]
3 years ago
9

PLZZ HELP Explain how you can tell the difference between a physical and a chemical change.

Physics
1 answer:
Alecsey [184]3 years ago
7 0

Answer:

You can tell the difference between physical and chemical change by looking at the change that takes place like whether it is a permanent or temporary change.

Explanation:

  • physical change is temporary change and chemical change is permanent change
  • the original shape of the object can be restored in physical change whereas the original composition of the object will be lost in chemical change
  • in physical change the change is reversible whereas he chemical change is irreversible and the object takes new shape and position in this change.
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A person wears a hearing aid that uniformly increases the sound level of all audible frequencies of sound by 28.1 dB. The hearin
ladessa [460]

Answer:

 I₂ = 2.13 x 10⁻⁸ W/m²

Explanation:

given,

increase in sound level = 28.1 dB

frequency of the sound = 250 Hz

intensity = 3.3 x 10⁻¹¹ W/m²

Intensity delivered = ?

the difference of intensity level is give as

\beta_2-\beta_1 = 10log(\dfrac{I_2}{I_o}) - 10log(\dfrac{I_1}{I_o})

\beta_2-\beta_1 = 10(log(\dfrac{I_2}{I_o}) -log(\dfrac{I_1}{I_o}))

\beta_2-\beta_1 = 10(log(\dfrac{I_2}{I_1})

28.1= 10(log(\dfrac{I_2}{I_1})

log\dfrac{I_2}{I_1}=2.81

\dfrac{I_2}{I_1}=10^{2.81}

 I₂ = 645.65 I₁

 I₂ = 645.65 x 3.3 x 10⁻¹¹

 I₂ = 2.13 x 10⁻⁸ W/m²

4 0
3 years ago
What happens to the saturation for when adding salt to water at room temperature
saw5 [17]
Saturated and Unsaturated Solutions
Table salt (NaCl) readily dissolves in water. Suppose that you have a beaker of water to which you add some salt, stirring until it dissolves. So you add more and that dissolves. You keep adding more and more salt, eventually reaching a point that no more of the salt will dissolve no matter how long or how vigorously you stir it. Why? On the molecular level, we know that action of the water causes the individual ions to break apart from the salt crystal and enter the solution, where they remain hydrated by water molecules. What also happens is that some of the dissolved ions collide back again with the crystal and remain there. Recrystallization is the process of dissolved solute returning to the solid state. At some point the rate at which the solid salt is dissolving becomes equal to the rate at which the dissolved solute is recrystallizing. When that point is reached, the total amount of dissolved salt remains unchanged. Solution equilibrium is the physical state described by the opposing processes of dissolution and recrystallization occurring at the same rate.
While this shows the change of state back and forth between solid and aqueous solution, the preferred equation also shows the dissociation that occurs as an ionic solid dissolves.



When the solution equilibrium point is reached and no more solute will dissolve, the solution is said to be saturated. A saturated solution is a solution that contains the maximum amount of solute that is capable of being dissolved. At 20°C, the maximum amount of NaCl that will dissolve in 100. g of water is 36.0 g. If any more NaCl is added past that point, it will not dissolve because the solution is saturated. What if more water is added to the solution instead? Now more NaCl would be capable of dissolving in the additional solvent. An unsaturated solution is a solution that contains less than the maximum amount of solute that is capable of being dissolved.

When 30.0 g of NaCl is added to 100 ml of water, it all dissolves, forming an unsaturated solution. When 40.0 g is added, 36.0 g dissolves and 4.0 g remains undissolved, forming a saturated solution.

How can you tell if a solution is saturated or unsaturated? If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.
5 0
3 years ago
Substance A has twice the specific heat capacity as substance B. If 1000 J of heat are added to 1.0 kg of each substance, compar
Mrrafil [7]
Substance A would have a delta T (change in temp) rise 1/2 the rise in substance B.

Q=mc x delta T

Q= heat energy in Joules
m= mass of substance heated or cooled
c= specific heat
ΔT is change in temp.

Solve for change in temp=. Q/mc

Specific heat and mass are not inversely proportional to heat energy input.

Putting into real world scenario of using water to heat a building.

Specific heat of water is 1.
It takes 1 btu to raise one pound of water 1 degF. at a base of 60 degF

Acetone specific heat is .51

So it takes half the amount of heat input to get a 100 degree ΔT, as compared to water.
4 0
3 years ago
What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

4 0
3 years ago
. If force (F), work (W) and velocity (v) are taken as fundamental quantities.
alex41 [277]

Answer:

∴ [T]=[WF−1V−1]

Hope this answer is right!!

7 0
2 years ago
Read 2 more answers
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