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3 years ago

Establishing a territory helps an animal in mating, raising its young, and finding food.

1 answer:

8 0

The correct answer is TRUE because when a territory is established its like a family purchasing its home, the family feels more comfortable in their home while an animal feels more comfortable in their territory

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Why is a solution not a pure substance

Firlakuza [10]

It is not a pure substance, because a solution are mixed chemicals in a way that the molecules are not bonded with one another. Thus, separating them from compounds and elements, which are pure substances.

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3 years ago

Explain how materials are classified on the basis of electrical conductivity. Give two uses of these materials in day to day lif

vitfil [10]

Answer: On the basis of electrical conductivity, materials are classified as conductors, semi-conductors and non-conductors. Conductors : Substances through which electricity can easily pass through are known as conductors. ... For example, glass, wood are non-conductors. Wood is used to make tables, desks etc.

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3 years ago

Kira dissolved different solutes in 100 ml of water. She recorded the amount of each solute that would dissolve at different tem

Radda [10]

This information confirms that temperature and solubility are related. The higher the temperature is, the more soluble.

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2 years ago

The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

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3 years ago

What intermolecular forces occur between two methane molecules?

nikklg [1K]

<span><span>Dipole-dipole interactions , example: ammoni </span><span>forces, example: methane, CH4</span><span>Hydrogen bonding example: water, H2O </span></span>

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3 years ago