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2 years ago

As you move upward, from level, in an energy pyramid, available energy ____.

Chemistry

1 answer:

svetlana [45]2 years ago

8 0

The available energy decreases as one moves upward in an energy pyramid.

<h3>Energy pyramid</h3>

The energy pyramid represents a model of how energy is transferred from one trophic level to another in ecosystems.

Energy is transferred from producers to primary consumers, from primary to secondary consumers, from secondary to tertiary consumers, and so on.

Only about 10% of the available energy in one trophic level is transferred to the next with the remaining 90% being lost as heat to the environment.

Thus, as one moves up the energy pyramid, the available energy decreases. This is why organisms at the higher end of the energy pyramid have to devise an efficient way of extracting energy from their foods.

More on energy pyramid can be found here: brainly.com/question/2515928

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A 100.0g sample of tin is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150. g of water at 2

Snowcat [4.5K]

Explanation:

It is known that specific heat of water is 4.184 $J/g^{o}C$ and atomic mass of tin is 118.7 g/mol. For the given situation,

$Q_{lost} = Q_{gained}$

Let us assume that,

$m_{1}$ = mass of Sn

$m_{2}$ = mass of $H_{2}O$

Therefore, heat energy expression for heat lost and gained is as follows.

$Q_{lost} = Q_{gained}$

$m_{1}C_{1}(T_{2} - T_{1}) = m_{2}C_{2}(T_{1} - T_{2})$

$100 g \times C_{1} \times (100^{o}C - 27.4^{o}C) = 150 g \times 4.184 /g^{o}C \times (27.4^{o}C - 25^{o}C)$

$7260C_{1} = 150 \times 4.184 \times 2.4$

$C_{1} = \frac{1506.24}{7260}$

= 0.207 $J/g^{o}C$

For, 118.7 g the specific heat of tin will be calculated as follows.

$C_{1} = 0.207 J/g^{o}C \times 118.7 g$

= 24.5 $J/mol^{o}C$

Thus, we can conclude that specific heat of tin is 24.5 $J/mol^{o}C$ .

3 0

3 years ago

The standard free energy change for a reaction can be calculated using the equation ΔG∘′=−nFΔE∘′ ΔG∘′=−nFΔE∘′ where nn is the nu

Lelechka [254]

Answer:

ΔG°′ = 1.737 KJ/mol

Explanation:

The reaction involves the transfer of two electrons in the form of hydride ions from reduced coenzyme Q, CoQH₂ to fumarae to form succinate and oxidized coenzyme Q, CoQ.

The overall equation of reaction is as follows:

fumarate²⁻ + CoQH₂ ↽⇀ succinate²⁻ + CoQ ; ΔE∘′=−0.009 V

Using the equation for standard free energy change; ΔG°′ = −nFΔE°′

where n = 2; F = 96.5 KJ.V⁻¹.mol⁻¹; ΔE°′ = 0.009 V

ΔG°′ = - 2 * 96.5 KJ.V⁻¹.mol⁻¹ * 0.009 V

ΔG°′ = 1.737 KJ/mol

6 0

3 years ago

What is the mass of a gold bar that is 7.379*10^-4 m^3 in volume?

NISA [10]

Mass to volume
M^3 --> cm^3
Cm^3 --> mL
Centi = 10*-2
7.379*10-4 / 1*10-2 = .07379

8 0

3 years ago

Calculate the energy required to ionize a ground state hydrogen atom. report your answer in kilojoules.

hjlf

First we find for the wavelength of the photon released due to change in energy level. We use the Rydberg equation:

1/ʎ = R [1/n1^2 – 1/n2^2]

where,

ʎ is the wavelength

R is the rydbergs constant = 1.097×10^7 m^-1

n1 is the 1st energy level = 1

n2 is the higher energy level = infinity, so 1/n2 = 0

Calculating for ʎ:

1/ʎ = 1.097×10^7 m^-1 * [1/1^2 – 0]

ʎ = 9.1158 x 10^-8 m

Then calculate the energy using Plancks equation:

E = hc/ʎ

where,

h is plancks constant = 6.626×10^−34 J s

c is speed of light = 3x10^8 m/s

E = (6.626×10^−34 J s * 3x10^8 m/s) / 9.1158 x 10^-8 m

E = 2.18 x 10^-18 J = 2.18 x 10^-21 kJ

This is still per atom, so multiply by Avogadros number = 6.022 x 10^23 atoms / mol:

E = (2.18 x 10^-21 kJ / atom) * (6.022 x 10^23 atoms / mol)

E = 1312 kJ/mol

3 0

3 years ago

A rectangular block has the following dimensions: 3.21 dm, 5.83 cm, and 1.84 in. The block has a mass of 1.94 kg. What is the de

spin [16.1K]

The density of the rectangular block in g/mL is 7.0.

Given the following data:

Mass of block = 22.8 gra1.94 kg

Length of block = 3.21 cm

Width of block = 5.83 cm

Height of block = 1.84 in.

To find the density of the block in g/mL:

First of all, we would determine the volume of the rectangular block by using the following formula:

$Volume = length$ × $width$ × $height$

Conversion:

1 in = 2.54 cm

5.83 in = X cm

Cross-multiplying, we have:

$X = 2.54(5.83)\\\\X = 14.81 \; cm$

$Volume = 3.21$ × $5.83$ × $14.81$

Volume = 277.16 cubic centimeters.

Note: Milliliter (mL) is the same as cubic centimeters.

1000 grams = 1 kg

Y grams = 1.94 kg

Cross-multiplying, we have:

Y = 1940 grams

Now, we can find the density:

$Density = \frac{Mass}{Volume}\\\\Density = \frac{1940}{277.16}$

Density = 7.0 g/mL

Therefore, the density of the rectangular block in g/mL is 7.0.

4 0

2 years ago