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hichkok12 [17]
3 years ago
7

Why is the base of an energy pyramid wider than the top

Chemistry
1 answer:
Zepler [3.9K]3 years ago
5 0

It is due to a lack of providable energy to the next trophic level of the energy pyramid. Primary consumers only obtain around 10% of energy that producers have, and the energy depletes as you move further and further up the food chain, until you reach the tertiary consumers which have the least amount of energy at their disposal, meaning there are less of them than other organisms further down the energy pyramid.


Hope this helped!

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The bond angles in SCl2 are expected to be Multiple Choice a little more than 109.5°. 109.5°. 120°. a little less than 109.5°. 1
nalin [4]

Answer:

a little less than 109.5°

Explanation:

SCl2 has four regions of electron density around the central atom of the molecule. This implies that it has a tetrahedral electron domain geometry with an expected bond angle of 109.5° according to valence shell electron pair repulsion theory.

However, there are two lone pair of electrons on the central atom of the molecule which decreases the bond angle a little less than 109.5° owing to repulsion between electron pairs.

4 0
2 years ago
PLEASE HELP!!!
neonofarm [45]

Answer:

Cl_2+2NaI\rightarrow I_2+2NaCl

Explanation:

Hello there!

In this case, according to the described chemical reaction, Cl2 replaces iodine in NaI in order to produce I2 and NaCl:

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It is possible to realize how chlorine replaces iodine in agreement with the single displacement reaction. Moreover, since chlorine and iodine atoms are not correctly balanced, we add a 2 in front of both NaI and NaCl in order to do so:

Cl_2+2NaI\rightarrow I_2+2NaCl

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8 0
2 years ago
The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.
Marina86 [1]

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

\frac{dA}{dt}= kA

\Rightarrow \frac{dA}{A}=kdt

Integrating both sides

ln A= kt+c₁

\Rightarrow A= e^{kt+c_1}

\Rightarrow A=Ce^{kt}         [e^{c_1}=C]

The initial condition is A(0)= A₀,

\therefore A_0=Ce^{0.k}

\Rightarrow C=A_0

\therefore A=A_0e^{kt}.........(1)

Given that the Pb_{209}  has half life of 3.3 hours.

For half life A=\frac12 A_0 putting this in equation (1)

\frac12A_0=A_0e^{k\times3.3}

\Rightarrow ln(\frac12)= 3.3k     [taking ln both sides, ln \ e^a=a]

\Rightarrow k=\frac{ln \frac12}{3.3}

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

\therefore 0.2=1e^{(-0.21)\times t}

\Rightarrow e^{-0.21t}=0.2

\Rightarrow -0.21t= ln(0. 2)

\Rightarrow t= \frac{ln (0.2)}{-0.21}

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

3 0
3 years ago
What is the concentration of the base (NaOH) in this titration?
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The answer is B for fact

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