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Mice21 [21]
3 years ago
13

An animal which eats plants and animals is a(n)

Chemistry
1 answer:
Vilka [71]3 years ago
6 0

Answer:

An Omnivore

Explanation: An <u>omnivore</u> is a kind of animal that eats either other animals or plants. Some omnivores will hunt and eat their food, like carnivores, eating herbivores and other omnivores. Some others are scavengers and will eat dead matter. Many will eat eggs from other animals.

Omnivores eat plants, but not all kinds of plants. Unlike herbivores, omnivores can't digest some of the substances in grains or other plants that do not produce fruit. They can eat fruits and vegetables, though. Some of the insect omnivores in this simulation are pollinators, which are very important to the life cycle of some kinds of plants.

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What are the 6 elements of wether
Citrus2011 [14]
The six basic elements of weather include temperature, humidity, atmospheric pressure, wind, precipitation, and cloudiness.

~I knew this was asked 5 days ago, but I hope this still helps.








8 0
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
To what Kelvin temperature must a balloon
snow_tiger [21]

Answer:

T_2=261.46\ K

Explanation:

It is given that,

Original temperature, T_1=323^{\circ}C=596.15\ K

Original volume, V_1=2.85\ L

We need to find the temperature if the volume of the balloon to be shrink to 1.25 L.

According to Charles law, at constant pressure, V\propto T

It would means, \dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}

T₂ = ?

T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{1.25\times 596.15}{2.85}\\\\T_2=261.46\ K

So, the new temperature is 261.46 K.

4 0
3 years ago
What role do electrons play in creating intermolecular forces
Alex73 [517]

repartExiplanation:cionde atomos

7 0
2 years ago
Read 2 more answers
When 20.0 grams of an unknown compound are dissolved in 500. grams of benzene, the freezing point of the resulting solution is 3
Lelu [443]

Answer: 120g/mol

Explanation:

The first step we are to take is to calculate the freezing point depression of the solution.

ΔT(f) = freezing point of pure solvent - freezing point of solution

ΔT(f) = 5.48 - 3.77

ΔT(f) = 1.71°C

Next we are to calculate the molal concentration of the solution using freezing point depression

ΔT(f) = K(f) * m

m = ΔT(f)/K(f)

m = 1.71/5.12

m = 0.333 molal

Now, we calculate the molecular weight of the unknown...

m = 0.333 mol = 0.333 mol X per kg of benzene

moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene

moles of X = 0.1665

molecular weight of X = 20g of X/0.1665

molecular weight of X = 120/mol

5 0
3 years ago
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