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Novosadov [1.4K]
3 years ago
9

A certain large pond had many plants growing in it. It also had a population of fish living in it. One day, a chemical spilled i

nto the water. This chemical caused the plants to die.
A scientist monitored the pond after the chemical spill. She found that the amount of dissolved oxygen in the pond gradually decreased. Once it reached about half of its original amount, the fish in the pond began to die.

Which statement best explains this information?

A. Fish produce oxygen through respiration, and this oxygen is used by plants.

B. All the oxygen in a pond comes directly from the atmosphere, so fish do not need plants.

C. Plants produce oxygen through photosynthesis, and fish depend on this oxygen to survive.

D. Fish and plants compete with each other for the dissolved oxygen in a pond.

Pls help!!
Chemistry
1 answer:
DaniilM [7]3 years ago
6 0

Answer:

Hi! your answer would be C. Plants produce oxygen through photosynthesis, and fish depend on this oxygen to survive.

Explanation:

I took the test and got this question right!

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What types of chemical bonds?​
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Make a list of silver current industrial uses. Try to include the approximate annual demand if possible. Create a table to displ
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What do you think is the primary reason why the Unionid mussel population declined whenever there is a dramatic increase in the
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The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper
LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

t_{1/2}=\frac{1}{k\times a_0}

k = rate constant =?

a_0 = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

a) Initial concentration of XY = a_o=0.100 M

Rate constant of the reaction = k = 6.96\times 10^{-3} M^{-1} s^{-1}

Half life of the reaction is:

t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}

=1,436.78 s

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = 6.20\times 10^{-2} M

\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become 6.20\times 10^{-2} M.

e)  Initial concentration of SO_2Cl_2= 0.050 M

Final concentration after time t = x

t = 55.0 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0
3 years ago
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