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A whale has a calf and a bacteria replicates through cloning. Both of these are examples of what?

Answers

Answer 1

Answer:

Either growth and development or reproduction

Explanation:

Answer 2

Answer:

r e p r o d u c t i o n

Explanation:


Related Questions

Plants need to take in gases and release gases during their
everyday life processes. What part allows plants to do this?

Answers

Answer:

The Stomata

Explanation:

The Stomata is found in the leaves of plants and controls their gas exchange with the environment around them.

help i need to pass this class

Answers

Answer: It would be B

Explanation: a magnet moves from north to south

due to pole attraction is more in the north region

what is a divalent union​

Answers

Answer:

. A divalent cation is an cation with valence of 2+. This type of ion may form two chemical bonds with anions.

Explanation:

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If 6.00 g of carbon is burned completely, what volume of carbon dioxide gas is released at STP? 6.0 L 11.2 L 22.4 L 134.4 L​

Answers

Answer:

11.2

Explanation:

Find molar mass of carbon 12.01 g/mol then divide by 6.00 g of Carbon, then divide by STP 22.4 L = 11.19 and round to sig fig = 11.2 L

11.2 L volume of carbon dioxide gas is released at STP. Hence, option B is correct.

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

For this question, use the formula PV=nRT firstly, define your knowns (this requires you to calculate moles of carbon)

Molar mass of carbon = 12.01 g/mol

Moles carbon=[tex]\frac{mass}{molar \;mass}[/tex]

Moles carbon=[tex]\frac{6.00 g}{12.01 g/mol }[/tex]

Moles carbon= 0.4995836803 moles

P=101.3 kPa

V=?

n= 6.0÷12 = 0.5

R= 8.31

T= 273 K

V= (0.5 x 8.31 x 273) ÷101.3    

=11.19

Therefore, 11.2 L of carbon dioxide will be produced.

Learn more about the ideal gas here:

https://brainly.com/question/27691721

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Which statement best explains why aluminum is a pure substance? Group of answer choices Aluminum is one type of atom and its composition is always the same. Aluminum has a composition that is evenly mixed. The aluminum atom cannot be seen with a microscope. Aluminum contains substances that are not bonded together.

Answers

Answer:  

The best explanation for both the first and second claims is that aluminum is a pure substance: Aluminum is one form of atom and its structure is always the same.  Aluminum has an evenly blended composition.

Explanation:

Aluminum is one type of atom and its composition is always the same, this statement also gives the property of a pure substance . In fact, pure substances are often homogeneous and contain only one form of atom or molecule.

One of a pure substance's key characteristics is that it has a uniform composition that is homologous in nature, and can be mixed uniformly throughout. ALUMINUM's second statement notes the property of a pure substance. Aluminum is also homologous in terms of occurrence.

The atoms are very small and have a diameter of around 1 x 10-10 meters. It's difficult to view them using a light microscope due to their limited size. Although it may not be possible to view an atom using a light microscope, a number of techniques have been developed, such as electron microscopy, to observe and research the structure of atoms. Therefore, the statement that the aluminum atom can not be seen through a microscope does not explain the pure content of aluminum.

This argument - 'Aluminum contains substances which are not bound together' does not explain the fact that it is a pure material as it notes that the property of a pure material does not exist in relation to its chemical bonds.

.Therefore, the first and second aluminum statements give the best description of a pure material.

Calculate the energy of a light having a wavelength of 6.80 x 102 nm.

Answers

Answer:

Energy, [tex]E=2.92\times 10^{-19}\ J[/tex]

Explanation:

Given that,

The wavelength of a light is [tex]6.8\times 10^2\ nm[/tex]

We need to find the energy of light. The formula for the energy of light is given by :

[tex]E=\dfrac{hc}{\lambda}\\\\E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{6.8\times 10^2\times 10^{-9}}\\\\E=2.92\times 10^{-19}\ J[/tex]

So, the energy of a light is [tex]2.92\times 10^{-19}\ J[/tex].

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