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2 years ago

If 100% is not passed from one trophic level to the next, where does this energy go?

Chemistry

2 answers:

Sonja [21]2 years ago

6 0

D is the answer to question

dangina [55]2 years ago

4 0

D) Organisms use their energy for all of these things and more

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What attracts heat more? black or aluminum foil?

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Answer:

"ALUMINUM FOIL" will conduct heat much more effectively than black construction paper.

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2 years ago

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Define an independent variable in an experiment. Explain the term with respect to the solubility of salt at different temperatur

nekit [7.7K]

Independent variable would be salt since you can't change it in this experiment.

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3 years ago

1) A light bulb takes in 30 of energy per second. It transfers 3j as use

natta225 [31]

Answer:

$\boxed{\text{10 \%}}$

Explanation:

The formula for efficiency is

$\begin{array}{rcl}\text{Efficiency} & = & \dfrac{\text{useful energy out}}{\text{energy in}} \times 100 \,\% \\\\\eta & = & \dfrac{w_{\text{out}}}{w_{\text{in}} } \times 100 \,\%\\\end{array}$

Data:

Useful energy = 3 J

Energy input = 30 J

Calculation:

$\begin{array}{rcl}\eta & = & \dfrac{\text{3 J}}{\text{30 J}} \times 100 \,\%\\\\\eta & = & 10 \, \%\\\end{array}\\\text{ The efficiency is }\boxed{\textbf{10 \%}}$

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2 years ago

Someone please help me dont scroll i need help

Diano4ka-milaya [45]

A line that stay s straight across the graph

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2 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

2 years ago

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