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Sever21 [200]
2 years ago
12

Which of the following organisms are carnivores?

Chemistry
1 answer:
sineoko [7]2 years ago
5 0

Answer:

Secondary consumers

Explanation:

Carnivores are organisms that only eat meat and that rely on others for food and energy; this means that they cannot produce food themselves.

Look at the answer choices:

- Decomposers: these are organisms that break down dead matter into material that can be used later by plants to grow; carnivores do not break down dead matter; rather, they eat meat.

- Primary consumers: these are usually herbivores; they are the second level of the energy pyramid, and they eat the organisms at the bottom level. The organisms at the bottom level are plants that make their own food. Obviously, plants are not meat, so carnivores are not primary consumers.

- Producers: these are the plants at the bottom of the energy pyramid that make their own food. Carnivores do not make their own food.

Thus, the answer is secondary consumers because this level of organisms eat the organisms at the primary consumer level.

Hope this helps!

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A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 417 pm. The diameter of the metal atom is:
enyata [817]

Answer:

b. 295 pm

Explanation:

To answer this question we need to use the equation of a face-centered cubic laticce:

Edge length = √8 R

<em>Where R is radius of the atom.</em>

<em />

Replacing:

417pm = √8 R

R = 147.4pm is the radius of the atom

As diameter = 2 radius.

Diameter of the metal atom is:

147.4pm* 2 =

295pm

Right answer is:

<h3>b. 295 pm </h3>

8 0
2 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N
Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

Now we have to calculate the dissociated concentration of N_2O_4.

The balanced equilibrium reaction is,

                             N_2O_4(g)\rightleftharpoons 2NO_2(aq)

Initial conc.           1.0 M          0

At eqm. conc.     (1.0-x) M    (2x) M

As we are given,

The percent of dissociation of N_2O_4 = \alpha = 28.0 %

So, the dissociate concentration of N_2O_4 = C\alpha=1.0M\times \frac{28.0}{100}=0.28M

The value of x = C\alpha = 0.28 M

Now we have to calculate the concentration of N_2O_4\text{ and }NO_2 at equilibrium.

Concentration of N_2O_4 = 1.0 - x  = 1.0 - 0.28 = 0.72 M

Concentration of NO_2 = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

Now put all the values in this expression, we get :

K_c=\frac{(0.56)^2}{0.72}=0.44

Therefore, the equilibrium constant K_c for the reaction is, 0.44

8 0
2 years ago
A deck of cards has dimensions 8.9cm x 5.72cm x 1.82 cm. What is the volume of the deck in cubic centimeters?
arsen [322]

Answer:

92.65256 cm^3

Explanation:

To find this, we can simply multiply all three dimensions to get the answer in cubic centimeters, and we get the answer above. If you want to be more specific, we can go by the sigfig rule and the answer would be rounded to 93 cm^3.

5 0
2 years ago
Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to
Delvig [45]

Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

7 0
2 years ago
Adding one proton to the nucleus of an atom
Sonbull [250]
Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number. So, adding or removing protons from the nucleus changes what element that atom is! For example, adding a proton to the nucleus of an atom of hydrogen creates an atom of helium.
7 0
2 years ago
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