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3 years ago

Which technique often results in the greatest habitat loss

Chemistry

2 answers:

tigry1 [53]3 years ago

7 0

Answer:

It is clear cutting

Explanation:

Clearcutting, clearfelling or clearcut logging is a forestry/logging practice in which most or all trees in an area are uniformly cut down.Clearcutting has been a controversial subject because of its environmental impact. While those who favour this method argue that it allows regenerating trees to have more access to sunlight while it reduces the risk of forest fires in some high-risk areas, opponents point out that in many cases, mass tree removal has caused soil erosion and the destruction of natural habitats for some animals.

Naddik [55]3 years ago

4 0

Answer: b. Clear cutting

Explanation:

Clear cutting is the correct option. Clear cutting involves the clearing up of entire vegetation cover of the region or cutting up of trees. This will lead to loss of canopy cover from a region. This will cause loss of biodiversity of a region as the animals, birds and insects will loose their habitat. Such faunal species may translocate to new region, decrease in number or may extinct.

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Which formula is an empirical formula?

poizon [28]

<h2>Hello!</h2>

The answer is:

The empirical formula is the option B. $NH_{3}$

The empirical formula of a compound is the simplest formula that can be written. On the opposite, the molecular formula involves a variant of the same compound, but it can be also simplified to an empirical formula.

$MolecularFormula=n(EmpiricalFormula)$

We are looking for a formula that cannot be simplified by dividing the number of molecules/atoms that conforms the compound.

Let's discard option by option in order to find which formula is an empirical formula (cannot be simplified)

A. $N_{2}O_{4}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$N_{2}O_{4}=2(NO_{2})$

B. $NH_{3}$

It's an empirical formula since it cannot be obtained by the multiplication of a whole number and the simplest formula. It's the simplest formula that we can find of the compound.

C. $C_{3}H_{6}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$C_{3}H_{6}=3(CH_{2})$

D. $P_{4}O_{10}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$P_{4}O_{10}=2(P_{2}O_{5})$

Hence, the empirical formula is the option B. $NH_{3}$

Have a nice day!

6 0

3 years ago

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What pressure will be exerted by 0.450 mol of a gas at 25°C if it is contained in a 0.650-L vessel?

iVinArrow [24]

Hey there!

The answer as well as the explanation is in the image attached. Let me know if there's anything you're unable to see.

Hope this helps!

5 0

3 years ago

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

Now put all the given values in the above formula, we get:

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0

4 years ago

Why does a solid change to a liquid when heat is added

Nata [24]

Particles will have more energy and will vibrate really fast.

(Hope this helps)

4 0

3 years ago

Read 2 more answers

Leave the answer blank if no precipitate will form. (Express your answer as a chemical formula.) Formula of precipitate ZnCl2(aq

Rasek [7]

Answer:

1. Zn(OH)₂ (s)

2. Ag₂CO₃ (s)

3. Ni₃(PO₄)₂(s)

4. No reaction

5. (NH₄)₂CO₃(s)

Explanation:

Let's state the equations and we analyse some solubility and precipitation information:

ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s) + 2KCl (aq)

All the salts from the halogens with group 1, are soluble.

The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble

K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)

All salts from nitrate are soluble

All salts from carbonates are insoluble

2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)

Salts from phosphates are insoluble

All salts from nitrate are soluble

NaCl(aq) + KNO3(aq) → NO REACTION

All salts from nitrate are soluble

All the salts from the halogens with group 1, are soluble

Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓

All salts from carbonates are insoluble

All the salts from the halogens with group 1, are soluble

7 0

3 years ago

Read 2 more answers