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Liono4ka [1.6K]
3 years ago
11

What factors could have affected your results?

Chemistry
1 answer:
AleksandrR [38]3 years ago
8 0

Answer:

See below  

Step-by-step explanation:

heat gained by metal + heat lost by water = 0

           m₁C₁ΔT₁           +         m₂C₂ΔT₂         = 0

C₁ = -(m₂C₂ΔT₂)/(m₁ΔT₁)

The factors determining C₁ are

  • mass of water
  • temperature change of water (T_f - Ti)
  • mass of metal
  • temperature change of metal (T_f - Ti)

Any factor that makes the numerator higher or the denominator lower than what you thought, will give a calculated C₁ that is too high (and vice versa).

The major sources of uncertainty are probably in determining the temperatures, especially the initial and final temperatures of the metal. However, you will have to decide what the principal factors were in your experiment.

For example, did the metal have a chance to cool during the transfer to the calorimeter? How easy was it to determine the equilibrium temperature, etc?

Factors Affecting the Calculation of Specific Heat Capacity

<u>          Too Low                 </u>              <u>      Too high                   </u>              

Water                                             Water

    Mass less than thought                Mass more than thought

    Ti lower                                         Ti higher

    T_f higher                                       T_f lower

Metal                                              Metal

    Mass more than thought             Mass less than thought

    Ti higher                                       Ti lower

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balandron [24]
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Hope this helps
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What is the theoretical yield of a reaction?
mestny [16]
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pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which beha
Fiesta28 [93]

Complete Question

Questions Diagram is attached below

Answer:

*  W=1142.86Joule

*  Q=997.7J

*  H=2140.5J

Explanation:

From the question we are told that:

Temperature T=337K

Pressure P=(60-55)Pa*10^5

VolumeV=(1.6-1.4)m^3*10^{-3}

Generally the equation for gas Constant is mathematically given by

\frac{P_2}{P_1}=\frac{V_1}{V_2}^n

 \frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n

 n=0.65

Therefore

Work-done

 W=\int{pdv}

 W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}

 W=1142.86Joule

Generally the equation for internal energy is mathematically given by

 Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}

 Q=997.7J

Therefore

 H=Q+W

 H=997.7J-11.42.9

 H=2140.5J

5 0
2 years ago
If a mass of a cube were 48.0g and its volume 24.0 cm3 what would its density be
EastWind [94]

Answer:

<h3>The answer is 2 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass = 48 g

volume = 24 cm³

We have

density =  \frac{48}{24}  \\

We have the final answer as

<h3>2 g/cm³</h3>

Hope this helps you

5 0
3 years ago
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