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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

You might be interested in

If 2.5L of solution is diluted to prepare 1.7L of a 0.8M solution, what was the original concentration?

mamaluj [8]

Answer:

0.544 M

Explanation:

First find the moles in the final solution

0.8 mols/L *1.7L

1.36 mols

so there is 1.36 mols in 2.5L

concentration will be 1.36/2.5

0.544 M

5 0

2 years ago

6.5 Moles of Al reacts with 7.2 Moles of H2O What is the limiting reactant and calculate the Theoretical Yield?

SVEN [57.7K]

Answer:

H₂O is the limiting reactant

Theoretical yield of 240 g Al₂O₃ and 14 g H₂

Explanation:

Find how many moles of one reactant is needed to completely react with the other.

6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O

We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.

Now find the theoretical yield:

7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃

7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂

Since the data was given to two significant figures, we must round our answer to two significant figures as well.

4 0

3 years ago

HELP ASAP

arlik [135]

Answer:

He's flying and was looking

Explanation:

6 0

3 years ago

10) Give two example of where you would find elements in our daily lives.

Phoenix [80]

Carbon is found in oil and gas.
Aluminum a light metal used in making pots and pans.
Bromine is used in photography.

6 0

3 years ago

Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to

vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

$H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)$

Next we will calculate how many moles of $H_2SO_4$ are present in 85.00 mL of 1.500 M sulfuric acid.

As, Molarity = $\frac{\text{moles of solute}}{\text{liters of solution
}}$

1.500 M = $\frac{n}{0.08500 L
}$

n = 0.1275 mol $H_2SO_4$

Now set up and solve a stoichiometric conversion from moles of $H_2SO_4$ to grams of $NaHCO_3$ . As, the molar mass of $NaHCO_3$ is 84.01 g/mol.

$0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})$

= 21.42 g $NaHCO_3$

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0

3 years ago