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2 years ago

Which consumers are cornivores

Chemistry

2 answers:

QveST [7]2 years ago

5 0

Answer:

Secondary consumers

Explanation:

hope this helps

Rus_ich [418]2 years ago

5 0

Answer:

Carnivores and omnivores are secondary consumers. Many carnivores eat herbivores. Some eat omnivores, and some eat other carnivores. Carnivores that consume other carnivores are called tertiary consumers.

E.g

Felines, ranging from domestic cats to lions, tigers, and other large predators.

Some canines, such as the Gray Wolf

Hyenas.

ferrets.

Polar Bears.

Pinnipeds ( seals, sea lions, walruses, etc.)

Birds of prey, including hawks, eagles, falcons, and owls.

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In the first 85.0 s of this reaction, the concentration of no dropped from 1.12 m to 0.520 m . calculate the average rate of the

goldenfox [79]

In the first 85.0 s of this reaction, the concentration of no dropped from 1.12 m to 0.520 m .

What is rate of a reaction?

The speed at which a chemical reaction takes place is the rate of the reaction. It is the concentration change per unit time of a reactant in a reaction.

Since the concentration of NO reduces to half its initial concentration in 85 seconds that is from 1.12m to 0.520m, it can be said that 85 seconds is the half life interval for the reaction, Hence on average, half reaction is completed in the time interval of 85 seconds.

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3 years ago

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

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3 years ago

Which of the following would violate laboratory safety rules? (1 point)

mixer [17]

Answer:

wearing open toed shoes

Explanation:

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