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Darya [45]
3 years ago
15

Please help its overdue

Chemistry
2 answers:
Paladinen [302]3 years ago
8 0

Answer:

First one:

The camel, reindeer and platypus does not belong. (Platypus live in the amazon river).

Second one:

The cactus does not belong and the spiky brown thing who looks like it is in the center of a desert does not belong too.

Third one:

Rainfall: The average annual precipitation of a temperate rainforest biome is 200cm. In much warmer areas, the average precipitation goes up to about 350 cm annually.

Temperatures:

In the file!!!

Other information:

<em>Temperate Forest </em>

Temperate forests are those found in the moderate climates between the tropics and boreal regions in both the Northern and Southern Hemisphere. They may also be called “four-season forests” because the midlatitude climates harboring them tend to experience four distinct seasons. A vast diversity of different forest types make up this broad category, from the broadly distributed temperate deciduous forests to pine woods and relatively geographically restricted temperate rainforests.  

Temperate forest often refers specifically to the temperate deciduous forests widespread in eastern North America and Eurasia, but other temperate-forest types exist in the planet's middle latitudes where moderate, frequently four-season climates encourage diverse tree growth.

<em>Locations and Climates</em>

Temperate forests range across large areas of North America and Eurasia as well as smaller portions of the Southern Hemisphere. Temperate deciduous forests, the “signature” temperate forest type, reach their greatest extent in the eastern United States and Canada, Europe, China, Japan and western Russia. Climatically speaking, temperate forests tend to experience fairly long growing seasons and decent amounts of rainfall that may be spread fairly evenly across the year or concentrated in a particular season; deciduous hardwoods, which lose their leaves in winter, dominate most major temperate forests. Drier temperate climates in, for example, western North America may see evergreen pines and other drought-tolerant conifers proliferate. Temperate rainforests, two-thirds of which lie in North America’s Pacific Northwest, experience milder, moister, often maritime-influenced climates than other temperate forests; those of the Pacific Northwest are unique in the dominance of conifers over hardwoods.

<em>Seasons in a Temperate Deciduous Forest</em>

During the winter, a temperate deciduous forest looks dead because leaves have fallen off most of the trees. Wildlife in these forests may endure the winter or migrate to warmer climates. Spring sees a rebirth of sorts with hardwoods leafing out and a proliferation of flowering shrubs and forbs. As days begin shortening and temperatures dropping in fall, the leaves of deciduous trees change color and begin dropping, while animals begin storing food for the winter and/or packing on body fat for winter survival or the energetic demands of migration.

<em>  </em>

<em>The Flora of Temperate Forests</em>

The soils of many temperate forests are fertile and support a rich diversity of trees. Temperate deciduous forests often feature varieties such as maples, oaks, elms and birches. Conifers such as pines and hemlocks may play a minority role in these hardwood-dominated communities, but, again, these needle-leaved trees may also form the majority in certain temperate ecosystems, such as the North American temperate rainforest and the pine forests of the southeastern U.S. A sub-variety of temperate forest found in so-called Mediterranean climates commonly features evergreen broadleaf trees, such as “live oaks” in California and parts of Southern Europe and eucalpts in Australia. Mosses, ferns and understory shrubs are common in many temperate forests.

<em>The Fauna of Temperate Forests</em>

<em>With their moderate climate and typically rich array of food resources, temperate forests tend to support a great diversity of wildlife. Koalas, possums, wombats and other marsupials roam Australian temperate forests, while in North American and Eurasian ecosystems deer, bears, foxes, wolves, squirrels and rabbits are common inhabitants. China’s temperate forests play host to giant and red pandas, which mostly eat bamboo. Many migratory songbirds nest in temperate forests, taking advantage of their spring and summer bounty of blossoms, berries, seeds and insects.</em>

Hoped I helped! Have a nice day!

djyliett [7]3 years ago
4 0
I hope this helps and the Platypus the reindeer and the camo don’t belong in the forest
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10 points. Please help.
ratelena [41]

Answer:

-191.7°C

Explanation:

P . V = n . R . T

That's the Ideal Gases Law. It can be useful to solve the question.

We replace data:

2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°

(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°

T° = 81.3 K

We convert T° from K to C°

81.3K - 273 = -191.7°C

6 0
3 years ago
Read 2 more answers
Question 231.15 pts A 0.200 M solution of a weak monoprotic acid HA is found to have a pH of 3.00 at room temperature. What is t
MariettaO [177]

Answer:

The correct answer is: Ka= 5.0 x 10⁻⁶

Explanation:

The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:

               HA               ⇄        H⁺        +          A⁻

t= 0      0.200 M                     0                     0

t              -x                             x                       x

t= eq      0.200M -x               x                       x

At equilibrium, we have the following ionization constant expression (Ka):

Ka= \frac{ [H^{+}]  [A^{-} ]}{ [HA]}

Ka= \frac{x x}{0.200 M -x}

Ka= \frac{x^{2} }{0.200 M - x}

From the definition of pH, we know that:

pH= - log  [H⁺]

In this case, [H⁺]= x, so:

pH= -log x

3.0= -log x

⇒x = 10⁻³

We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:

Ka= \frac{(10^{-3})^{2}  }{0.200 - (10^{-3}) }= \frac{10^{-6} }{0.199}= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶

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How many atoms of phosphorus are in 3.80 mol of copper(II) phosphate?
Sladkaya [172]
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An<br> a transition metal with 91 protons and electrons?
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3 years ago
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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
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