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loris [4]
2 years ago
10

Organisms are classified into domains and other groups by similar characteristics. There are three domains: Archaea, Bacteria, a

nd Eukarya. Which of the following characteristics is used to classify organisms into domains?
genetic similarities


similarities in habitat


presence or absence of a nucleus


presence or absence of photosynthesis
Chemistry
1 answer:
Anton [14]2 years ago
8 0

Answer:

presence or absence of a nucleus

Explanation:

These classification of organisms into broad domains is based on the present or absence of nucleus in the cell of an organism.

The archaea are prokaryotes and they lack a distinct cellular nuclei.

Bacteria are similar to archaea but bacteria have only one RNA polymerase

Eukarya have true nucleus and membrane bound organelles .

So, the variations in their cells are used to classify organisms into the broad categories

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Physical chemistry is the area that deals with the mechanisms, the rate, and the energy transfer that occurs when matter undergoes a change. So the correct option for the given question is option “B”. Whenever there is a change in the state of matter, the physical properties definitely undergo a change. A solid matter will always have a definite shape and volume, while a liquid has a definite volume but no shape. Gases do not have either definite volume or shape. Chemical properties might not be affected, but physical properties are definitely impacted when a matter undergoes change.



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2 years ago
When iridium-192 is used in cancer treatment, a small cylindrical piece of 192ir, 0.6 mm in diameter and 3.5 mm long, is surgica
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Iridium-192 is used in cancer treatment, a small cylindrical piece of 192 Ir, 0.6 mm in diameter (0.3mm radius) and 3.5 mm long, is surgically inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?

Let us start by computing for the volume of the cylinder. V = π(r^2)*h where r and h are the radius and height of the cylinder, respectively. Let's convert all given dimensions to cm first. Radius = 0.03 cm, height is 0.35cm long.

V = π * (0.03cm)^2 * 0.35 cm = 9.896*10^-4 cm^3

Now we have the volume of 192-Ir, let's use the density provided to get it's mass, and once we have the mass let's use the molar mass to get the amount of moles. After getting the amount of moles, we use Avogadro's number to convert moles into number of atoms. See the calculation below and see if all units "cancel":

9.896*10^-4 cm^3 * (22.42 g/cm3) * (1 mole / 191.963 g) * (6.022x10^23 atoms /mole)

= 6.96 x 10^19 atoms of Ir-122 are present.

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