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2 years ago

What are some non examples of biodiversity

2 answers:

8 0

A monocrop is a non example of biodiversity because it contains only one species, such as all corn, therefore there is very little biodiversity.

777dan777 [17]2 years ago

7 0

Here is the answer
There are multiple ways to look at the concept of biodiversity.
There is diversity on the genetic level within populations, diversity within ecosystems among the species that inhabit them, and even diversity among ecosystems.
<span>You can also look at different scales of biodiversity. For example you can look at a square meter, and see 20 species in it and say that's a pretty diverse square meter. But then, if you look at the neighboring square meter and see the same 20 species, you might begin to question how diverse this ecosystem really is. In contrast, you can look at a square meter of another ecosystem and see 15 species, while its neighboring square meter has a completely different set of 15 species. Which is the ecosystem with greater biodiversity? It all depends on the scale you are interested in.</span>

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Identify the type of solid from the description. Select from the following choices: metallic solid, network solid, molecular sol

scZoUnD [109]

First of all, there are five types of solid materials:

Metallic solids which are solids composed of metal atoms that are held together by metallic bonds.

Network solid is a chemical compound in which the atoms are bonded by covalent bonds in a continuous network extending throughout the material.

Molecular solid is a solid consisting of discrete molecules.

Ionic solid is a chemical compound composed of ions held together by electrostatic forces termed ionic bonding.

Amorphous solid is non-crystalline solid that lacks the long-range order that is characteristic of a crystal.

Now, after the defined all the types of solid materials in the equation lets to solve it.

A. the answer is the network solids, because covalent bonds are relatively strong, covalent are typically characterized by hardness, strength, and high melting points.

B. the answer is the metallic solids, due to that heat conduction occurs when a substance is heated and the particles will gain more energy vibrating more. These molecules then bump into nearby particles and transfer some of their energy to them and in metals this process have a higher probability than in the case of other solids due to the nature of the chemical bonds. It also has a range of hardness due to the strength of metallic bonds which varies dramatically.

C. the answer is the ionic solid; due to positive and negative ions which are bonded to form a crystalline solid held together by charge attractions.

4 0

2 years ago

As photosynthesis occurs in chloroplasts, O2 is produced from _____ via a series of reactions associated with _____. View Availa

Monica [59]

Answer:

H₂O, Photosystem II

Explanation:

Photosynthesis is the process that enables autotrophs such as green plants and algae to generate food using water, carbon dioxide and energy from the sun.
It occurs in two phases, that is, the light-dependent phase and the light-independent phase.
During photosystem II energy from the sun is used to break-down water molecules to yield oxygen and hydrogen ions. Oxygen is released away to the atmosphere while hydrogen ions are used in the next phase to generate ATP molecules.

6 0

2 years ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

1 year ago

How much heat is required to convert 422 g of liquid h2o at 23.5 °c into steam at 150 °c?

Misha Larkins [42]

Heat is given by multiplying the specific heat capacity of a substance by mass and the change in temperature. The heat capacity of water is Approximately 4184 J/K/C.
Therefore, heat = mc0 mass in kg
= (422/1000) × 4184 × (100-23.5)
= 135072.072 J
Latent heat of vaporization is 2260 kJ/kg
Thus the heat will be 0.422 × 2260000 = 953720 J
Heat to raise steam from 100 to 150
2000 × 0.422 ×50 = 42200 J
Thus the heat required is (135072.072 + 953720 + 42200) = 1330992.07 Joules or 1330 kilo joules

4 0

1 year ago

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Citric acid (h3c6h5o7) is a product of the fermentation of sucrose (c12h22o11) in air. determine the mass of citric acid produce

Alex

2.50 x 2/1 = 5 mol of Citric Acid
5 x (3+72+5+112) = 960g of Citric Acid

Answer: 960g of Citric Acid

6 0

2 years ago

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