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3 years ago

What type of consumer eats only producers?

2 answers:

8 0

Answer: Primary consumers make up the second trophic level. They are also called herbivores. They eat primary producers—plants or algae—and nothing else. For example, a grasshopper living in the Everglades is a primary consumer

FrozenT [24]3 years ago

5 0

Answer:

primary consumer make up the second trophic level .they are also herbivores they eat primary consumer plants or alger and nothing else .for example a grasshopper living in the everglades is a primary consumer

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Which part of the factory is most similar to the nucleus of a living cell

labwork [276]

The control room is similar to the nucleus, because the nucleus is in charge of cell functions.

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3 years ago

Which correctly summarizes the trend in electron affinity?

ra1l [238]

The answer is it tends to be more negative down a group. This is because as you go down the periodic table, the elements have more electron shells in their atoms. This makes the outermost shells less attracted to the nucleus due to their greater distances from the nucleus. Therefore, these shells are less likely to attract electrons (hence lower electron affinity) and are even more likely to lose electrons from their outer electron orbits.

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4 years ago

Read 2 more answers

If a reaction mixture initially contains a co concentration of 0.1470 and a cl2 concentration of 0.175 at 1000k. what is the equ

NikAS [45]

When we have this equation:
CO(g) + Cl2(g) ↔ COCl2(g)
intial 0.147 0.175 0
change -X -X +X
final (0.147-X) (0.175-X) X

so from the ICE table, we substitute in Kc formula :(when we have Kc = 255)

Kc = [COCl2]/[CO][Cl2]
255= X / (0.147-X)(0.175-X)
255 = X / (X^2 - 0.322 X + 0.025725)
X = 0.13
∴[CO] = 0.147 - X = 0.147 - 0.13
= 0.017 m

3 0

3 years ago

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilo

Lynna [10]

Answer:

NA = 6.8 E-12 Kg H2(g) / hour

Explanation:

steady-state diffusion of A through non-diffuser B:

NA = (DAB/RTz)(p*A1 - p*A2)

∴ (A): H2(g)

∴ (B): Pd

∴ DAB = 1.7 E-8 m²/s

∴ p*A1 = 2.0 Kg H2 / m³ Pd

∴ p*A2 = 0.4 Kg H2 / m³ Pd

∴ z = 6 mm = 6 E-3 m

∴ T = 600°C ≅ 873 K

∴ R = 8.314 J/mol.K = 8.314 N.m/mol.K

⇒ NA = ((1.7 E-8)/(8.314)(873)(6 E-3))(2.0 - 0.4)

⇒ NA = 6.246 E-10 mol/s.m³

for A = 0.25 m²

⇒ volume (v) = A×z = (0.25)(6 E-3) = 1.5 E-3 m³

∴ Mw H2(g) = 2.016 g/mol

⇒ NA = (6.246 E-10 mol/s.m³)(1.5 E-3 m³)(2.016 g/mol)(Kg/1000 g)(3600 s/h)

⇒ NA = 6.8 E-12 Kg H2(g)/h

4 0

3 years ago

Forty miles above the earth's surface the temperature is 250k and the pressure is only 0.20 torr. what is the density of air at

alexira [117]

For a Forty miles above the earth's surface, the temperature at 250k and the pressure at only 0.20 torr, the density of air at this altitude is mathematically given as

d=3.732*10{-4}g/l

<h3>What is the density of air at this altitude?</h3>

Generally, the equation for the ideal gas is mathematically given as

PV=(m/mw)RT

Therefore

d=m/v

d=26,664*20/83.14*250

d=0.37152

d=3.732*10{-4}g/l

In conclusion, the density

d=3.732*10{-4}g/l