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2 years ago

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Temperate deciduous trees lose their leaves in Fall. Explain why trees in temperate rainforest and tropical rainforest don’t los

e their leaves. PLEASE HELP!!!!!!!!!!!!! 20 MORE MINUTES UNTIL I HAVE TO HAND THIS INN

1 answer:

andrey2020 [161]2 years ago

6 0

Answer:

So trees in temperate don't lose their leaves because the weather events aren't harsh enough.

Trees in tropical rainforest don't lose their leaves because they are a different type of tree known as evergreens that are green all year round.

Explanation:

Ok so first we'll define some things

Deciduous Trees= Trees that lose all of their leaves for part of the year.

Trees shed their leaves trees to try and survive harsh weather events.

Temperate deciduous trees lose their leaves in fall to better survive the winter conditions of extreme cold and reduced daylight.

Temperate rainforests = An area that doesn't experience extremely cold or extremely hot temperatures or what we would call harsh weather events.

Broad-leaved trees in tropical rainforests are known evergreen, they are known as this as they are green all year round.

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Enter the correct ground-state (or lowest energy) configuration based on the number of electrons: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10

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Answer:

FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.

Element Atomic number Electron configuration

sulfur 16 1s22s22p63s23p4

chlorine 17 1s22s22p63s23p5

argon 18 1s22s22p63s23p6

Explanation:

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2 years ago

Read 2 more answers

An imaginary element with BCC structure and has an atomic radius of 0.17 nm, with a molar mass of 56.08 g/mol. What is the densi

Oksi-84 [34.3K]

<u>Answer:</u> The density of the given element is $3.07g/cm^3$

<u>Explanation:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = 0.17 nm

a = edge length = ?

Putting values in above equation, we get:

$0.17=\frac{\sqrt{3}\times a}{4}\\\\a=\frac{0.17\times 4}{\sqrt{3}}=0.393nm$

To calculate the density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal = 56.08 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = $0.393nm=3.93\times 10^{-8}cm$ (Conversion factor: $1cm=10^{7}nm$ )

Putting values in above equation, we get:

$\rho=\frac{2\times 56.08}{6.022\times 10^{23}\times (3.93\times 10^{-8})^3}\\\\\rho=3.07g/cm^3$

Hence, the density of the given element is $3.07g/cm^3$

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2 years ago

For the net reaction: 2AB + 2C → A2 + 2BC, the following slow first steps have been proposed. AB → A + B 2C + AB → AC + BC 2AB →

puteri [66]

This is an incomplete question, here is a complete question.

For the net reaction: 2AB + 2C → A2 + 2BC, the following slow first steps have been proposed.

(1) AB → A + B

(2) 2C + AB → AC + BC

(3) 2AB → A₂ + 2B

(4) C + AB → BC + A

What rate law is predicted by each of these steps?

Answer : The rate law expression for the following reactions are:

(1) $\text{Rate}=k[AB]$

(2) $\text{Rate}=k[C]^2[AB]$

(3) $\text{Rate}=k[AB]^2$

(4) $\text{Rate}=k[C][AB]$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The general reaction is:

$A+B\rightarrow C+D$

The general rate law expression for the reaction is:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

R = rate law

k = rate constant

$[A]$ and $[B]$ = concentration of A and B reactant

Now we have to determine the rate law for the given reaction.

(1) The balanced equations will be:

$AB\rightarrow A+B$

In this reaction, $AB$ is the reactant.

The rate law expression for the reaction is:

$\text{Rate}=k[AB]$

(2) The balanced equations will be:

$2C+AB\rightarrow AC+BC$

In this reaction, $C$ and $AB$ are the reactants.

The rate law expression for the reaction is:

$\text{Rate}=k[C]^2[AB]$

(3) The balanced equations will be:

$2AB\rightarrow A_2+2B$

In this reaction, $AB$ is the reactant.

The rate law expression for the reaction is:

$\text{Rate}=k[AB]^2$

(4) The balanced equations will be:

$C+AB\rightarrow BC+A$

In this reaction, $C$ and $AB$ are the reactants.

The rate law expression for the reaction is:

$\text{Rate}=k[C][AB]$

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2 years ago

How many grams of Cu(OH)2 will precipitate when excess NaOH solution is added to 46.0 mL of 0.584 M CuSO4

Slav-nsk [51]

<h3>Answer:</h3>

2.624 g

<h3>Explanation:</h3>

The equation for the reaction is given as;

CuSO₄(aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
Volume of CuSO₄ as 46.0 mL;
Molarity of CuSO₄ as 0.584 M

We are required to calculate the mass of Cu(OH)₂ precipitated

We are going to use the following steps;

<h3>Step 1: Calculate the number of moles of CuSO₄ used</h3>

Molarity = Number of moles ÷ Volume

To get the number of moles;

Moles = Molarity × volume

= 0.584 M × 0.046 L

= 0.0269 moles

<h3>Step 2: Calculate the number of moles of Cu(OH)₂ produced </h3>

From the equation 1 mole of CuSO₄ reacts to give out 1 mole of Cu(OH)₂
Therefore; Mole ratio of CuSO₄ to Cu(OH)₂ is 1 : 1.

Thus, Moles of CuSO₄ = Moles of Cu(OH)₂

Hence, moles of Cu(OH)₂ = 0.0269 moles

<h3>Step 3: Calculate the mass of Cu(OH)₂</h3>

To get mass we multiply the number of moles with the molar mass.

Mass = Moles × Molar mass

Molar mass of Cu(OH)₂ is 97.561 g/mol

Therefore;

Mass of Cu(OH)₂ = 0.0269 moles × 97.561 g/mol

= 2.624 g

Thus, the mass of Cu(OH)₂ that will precipitate is 2.624 g

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2 years ago

Number 58!! PLEASE EXPLAIN!!! And show work

irina1246 [14]

A

and it is D for your ansower

Explanation:

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2 years ago