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mart [117]
2 years ago
15

How does the appearance of the protist differ from that of the onion sample?

Chemistry
1 answer:
lana [24]2 years ago
8 0

The appearance of the protist differ from that of the onion sample due to the presence of motile structures.

<h3>What is a Protist?</h3>

This is an eukaryote which could be unicellular or multicellular and examples include protozoa etc.

The major difference between a protist and onion sample is that protists have motile structures such as flagella, cilia etc while plant cells such as onions don't have.

Read more about Protist here brainly.com/question/2169979

#SPJ1

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EASY QUICK CHEM QUESTION!! Which line indicates a higher reaction rate?
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Calculate the molarity of 0.289 moles of FeC13 dissolved in 120 ml of solution
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The formula C2H4 can be described as
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Text 8.7 Using the virial equation of state for hydrogen at 298 K given in problem 7 (text 8.6), calculate a. The fugacity of hy
gregori [183]

Answer:

a). P = 688 atm

b). P = 1083.04 atm

c).Δ G = 16.188 J/mol    

Explanation:

a). Fugacity 'f' can be calculated from the following equations :

$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$

where, P = pressure ,  Z = compressibility

Now, the virial equation is :

$PV=R(1+6.4\times 10^{-4} P)$  ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

  $Z=1+6.4\times 10^{-4} P$

So from the fugacity equation ,

$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

$f=Pe^{6.4 \times 10^{-4} P}$

Putting the value of P = 500 atm in the above equation, we get,

f = 688 atm

b). Given f = 2P

    $2P=PE^{6.4 \times 10^{-4}P}$

   $2=E^{6.4 \times 10^{-4}P}$

  $\ln 2 = 6.4 \times 10^{-4}P$

  ∴  P = 1083.04 atm

c). dG = Vdp -S dt  at constant temperature, dT = 0

Therefore, dG = V dp

$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp $

            $=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$

           $=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$

$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$

Δ G = 16.188 J/mol  

7 0
3 years ago
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