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LiRa [457]
3 years ago
11

a valueable preserved biological specimen is weighed by suspeding it from a spring scale. it weighs 0.45 N when it is suspendedi

n air and 0.081 N when it is suspended in a bottle of alchol what is its dencity?
Engineering
1 answer:
Dmitry_Shevchenko [17]3 years ago
4 0

Answer:

ρ=962.16kg/m^3

Explanation:

The first thing we must do to solve is to find the mass of the specimen using the weight equation

w = mg

m=w/g

m=0.45/9.81=0.04587kg

To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).

We must bear in mind that the principle of archimedes indicates that the buoyant force is given by

F = ρgV

where V is the specimen volume and  ρ is the density of alcohol = 789kg / m ^ 3

considering the above we have the following equation

0.081=0.45-(789)(9.81m/s^2)V

solving for V

V=(0.081-0.45)/(-789x9.81)

V=4.7673x10^-5m^3

finally we found the density

ρ=m/v

ρ=0.04587kg/4.7673x10^-5m^3

ρ=962.16kg/m^3

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Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

<u>a) calculate the conversion exiting the first reactor </u>

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Given that there are two stream one  contains 16.1 mol/dm^3 while the other contains   0.9 wt% catalyst

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next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = \frac{Xai}{1-Xai}  ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm  ---  ( 2 )

<em>where : K = 0.311 ,  Im = 6.97   ( input values into equation 2 )</em>

Xai = 0.684

<u>B) calculate the conversion exiting the second reactor</u>

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

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To determine the value of the conversion exiting the second reactor  ( Xa2 ) we will use the relation below

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XA2 = 0.90

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Independent auto lots usually have <u>higher</u> finance rates than dealerships

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