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OlgaM077 [116]
2 years ago
9

In any food web, the organisms that are responsible for converting raw energy into usable chemical energy are collectively calle

d ________, while organisms
that recycle the nutrients trapped in dead organisms are collectively called ________.
A) producers; consumers.
B) producers; decomposers.
C) heterotrophs; consumers.
D) heterotrophs; autotrophs.
Chemistry
2 answers:
Elan Coil [88]2 years ago
8 0

Answer:

B. Producers and decomposers

Explanation:

Producers aka plants convert raw energy into chemical energy.

Decomposers are responsible for decomposing dead organisms.

Ivahew [28]2 years ago
5 0

Answer:

the answer is b

Explanation:

Because it provides for the restoration of the life cycle

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1.12 x 10^22 atoms to grams
Pie

Answer:

3.40

⋅

10

22

atoms

H

e

Explanation:

6 0
2 years ago
A mixture of oxygen, hydrogen, and nitrogen exerts a total pressure of 378 kPa. If the partial pressures of oxygen and hydrogen
sertanlavr [38]
What's the relationship between total and partial pressure? The total pressure is the sum of the parcial pressures!


So for us, it would be:

378= 212+101+x

where x is the parcial pressure of nitrogen.

Now we count:
378= 212+101+x
378=313+x
378-313=x
65=x

So the parcial pressure exerted by nitrogen is 65!

8 0
3 years ago
A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad
ASHA 777 [7]

0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

  • E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ} and
  • m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}

Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

  • Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ} (converted to kilojoules)
  • Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}
  • Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

  • E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})
  • E(\text{released}) =  Q(\text{lead})

Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0
3 years ago
Which organelles surround the cell?
GalinKa [24]

Answer:

Cell Wall, Cell Membrane, Nucleus, Endoplasmic Reticulum (ER), and Ribosome.

Explanation:

Hopefully this helps :).

5 0
3 years ago
Read 2 more answers
How many grams of NH3 can be produced from 12.0g of H2?
RSB [31]

Answer:

Balanced reaction:

3 H2 (g)  + N2 (g)  → 2 NH3 (g)

Use stoichiometry to convert g of H2 to g of NH3.  The process would be:

g H2 → mol H2 → mol NH3 → g NH3

12.0 g H2 x (1 mol H2 / 2.02 g H2) x (2 mol NH3 / 3 mol H2) x (17.03 g NH3 / 1 mol NH3) = 67.4 g NH3

Explanation: See above

Hope this helps, friend.

8 0
2 years ago
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