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3 years ago

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Which best describes the role of a primary consumer in a food web? Carnivore that feeds on producers Herbivore that feeds on pro

ducers Herbivore that feeds on primary consumers Carnivore that feeds on primary consumers

1 answer:

lina2011 [118]3 years ago

6 0

A primary consumer is a herbivore that feeds on producers. They are the first in the food chain to eat making them the primary consumers.

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Students rolled a glass marble and a baseball down an inclined plane, as shown below, and made observations.

Alja [10]

The answer is b i just took the quiz and got it right|| k12.gca student

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3 years ago

Read 2 more answers

Which one is the least reactive element is<br> Ar<br> Na<br> СІ<br> Mg

34kurt

Answer:

Ar

Explanation:

does not react with anything it has a full electron shell so it does not combine and it is a noble gas

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2 years ago

How do you describe the milk?

Lady bird [3.3K]

Answer:

emulsion

Explanation:

An emulsion is a mixture of two or more liquids that are normally immiscible. Emulsions are part of a more general class of two-phase systems of matter called colloids.

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3 years ago

How does changing the concentration of the reactants change the parts of a rate law? It changes the rate, R. It changes the rate

Sidana [21]

Answer

A. It changes the rate, R

Explanation

When we change the concentration of the reactants in a chemical reaction, it affects the rate of reaction that happens in the process. Typically, the rate of reaction will decrease with time if the concentration of the reactants decreases because the reactants will be converted to products. Similarly, the rate of reaction will increase when the concentration of reactants are increased.

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3 years ago

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

<u>Answer:</u> The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

<u>Initial:</u> 0.025 0.025

<u>At eqllm:</u> 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago