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UkoKoshka [18]
2 years ago
5

Which of these events would occur last in a food chain?

Chemistry
1 answer:
PSYCHO15rus [73]2 years ago
4 0
D. An animal decaying after it dies seem to be the right answer hopefully.
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The highest point is the crest
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What happens to water during evaporation?
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D would be your best bet because evaporation occurs when water is heated, it then vibrates and then magic!
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RATE LAW QUESTION !
vivado [14]
In general, we have this rate law express.:

\mathrm{Rate} = k \cdot [A]^x [B]^y
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4)  by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation \Delta \mathrm{Rate} = \Delta [B]^y

3.09 = (3)^y \implies y \approx 1

To this point, \mathrm{Rate} = k \cdot [A]^x [B]^1

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for \Delta \mathrm{Rate} = \Delta [A]^x

4=  (2)^x \implies x = 2

the rate law is

Rate = k·[A]²[B]
6 0
3 years ago
A ridged steel tank filled with 62.7L of nitrogen gas of 85.0L atm
mylen [45]

The final gas pressure : 175.53 atm

<h3>Further explanation</h3>

Maybe the complete question is like this :

A ridged steel tank filled with 62.7 l of nitrogen gas at 85.0 atm and 19 °C is heated to 330 °C while the volume remains constant. what is the final gas pressure?

The volume remains constant⇒Gay Lussac's Law  

<em>When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature  </em>

\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

P₁=85 atm

T₁=19+273=292 K

T₂=330+273=603 K

\tt P_2=\dfrac{P_1\times T_2}{T_1}\\\\P_2=\dfrac{85\times 603}{292}\\\\P_2=175.53~atm

8 0
2 years ago
If pollution is increased, then the cricket frog population will (increase/decrease/stay the same) over the span of five years.
stepladder [879]
I am guessing it will decrease
4 0
3 years ago
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