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Mariulka [41]
3 years ago
8

Myth: A dead organism is the same as a nonliving thing in science. Fact: Evidence:

Chemistry
2 answers:
pashok25 [27]3 years ago
6 0
Hello! :)

Myth: A dead organism is the same as a nonliving thing in science.

Fact: A non-living thing can be very different than a living thing. 

Evidence: Once- living things are made up different things then non-living things. While a rock is a nonliving thing, it doesn't contain the same build as living things. If we were to look at a dead once living being, we could probably conclude that it was a living thing since it contains things like cells, proteins, etc. That other thing, like dirt, wouldn't have. 

~'Manda
Ilya [14]3 years ago
4 0

I recently did this assignment!

Instructions: Read each myth (untruth). Reword it to make a factual statement. Then, give two to three reasons why the myth is untrue. Use complete sentences and support your answer with evidence, using your own words.  

________________________________________

Answer:

Myth: A dead organism is the same as a nonliving thing in science.

o Fact: In science, dead is the same as nonliving.

o Evidence: Things that are nonliving never had the characteristics of life, and never will. Things that are dead once did have the characteristics of life, but when they die, they lose some of the characteristics. That is why dead and non-living are NOT the same thing.

Hope this helped!

Have an Amazing day!

~Lola

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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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