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2 years ago

Which of the following is most helpful when identifying organism using a dichotomous key?

Chemistry

1 answer:

Anastaziya [24]2 years ago

3 0

Answer:

C. Its physical structures

Explanation:

A dichotomous key is defined as a tool used to differentiate between objects and organisms. A dichotomous key consist of series of question related to the species and at the end gives the scientific name of the species.

The characteristic that is most helpful when identifying organism using a dichotomous key is "Its physical structures". for example: if a person want to identify a tree, a dichotomous key ask about the physical structures such as if the tree has needles or leaves.

So, physical structures are most helpful while using a dichotomous key.

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Object A: has a mass of 2g a volume of .5L

malfutka [58]

An object will be floating in water if the density is lower and will be sink if the density is higher. Water density would be 1000g/liter or 1g/ml. To determine whether the object float or sink you need to calculate both object density.

The density of object A would be: 2g/ 0.5liter= 4g/liter
The density of object B would be: 3g/ 5liter= 0.6g/liter

It is clear that the object is much lighter than water so both objects would float.

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3 years ago

What’s the correct answer !!

Sati [7]

Answer:

From my side the correct option is "A"

Explanation: Reasons for its correction are ,

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3 years ago

Which of these would most likely be considered a pure substance?

Karolina [17]

Answer:

c. Water Vapour

Hope it helps.

5 0

3 years ago

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Why does Tom Zambrano think that renewable energy science is a good choice for a career?

ella [17]

Answer:

There will many opportunities in the field of renewable energy science and engineering. It is an exciting field, with a lot of variety and room for new ideas.

Explanation:

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3 years ago

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Calculate the amount of carbon dioxid gas in 1.505x10^23 molecules of the gas.

Daniel [21]

Explanation:

We need to find the amount of carbon dioxide gas in $1.505\times 10^{23}$ molecules of the gas.

We know that, 1 mole weighs 44 gram of carbon dioxide which contains $6.022\times 10^{23}$ number of molecules. It means that, $6.022\times 10^{23}$ number of molecules present in 44 grams of carbon dioxide molecule. So, $1.505\times 10^{23}$ number of molecules present in :

$=\dfrac{1.0505\times 10^{23}}{6.022\times 10^{23}}\times 44\\\\=7.675\ \text{grams}$

Hence, 7.675 grams of carbon dioxide is present in $1.505\times 10^{23}$ molecules of the gas.

4 0

3 years ago