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Zolol [24]
3 years ago
15

What are some geographic features that could be found in the hydrosphere?

Chemistry
1 answer:
nadezda [96]3 years ago
6 0

Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

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What was the name for the secret code used by many alchemists?
ahrayia [7]

Answer:

A. Zodiac

B. Palingenesis

C. Palabra mysteria

D. Decknamen

The correct answer is D. Decknamen.

Explanation:

5 0
3 years ago
Which answer below correctly identifies the type of change and the explanation for the boiling of water?
Feliz [49]

Answer:

physical change because even though gas formation was observed the water was undergoing a state change which means that it's original properties are preserved

5 0
3 years ago
HELP ASAP
Bezzdna [24]
A. I think sorry if it’s wrong
5 0
3 years ago
1s22s22p63s23p6 How many unpaired electrons are in the atom represented by the electron configuration above?
pickupchik [31]

Answer:

0

Explanation:

There are no unpaired electrons in the given element. It must be noted that for the atom above, we have even numbered electrons. The total electron we are having here is 18.

Now, we must also know that while the s orbital is not degenerate, the P orbital is degenerate. What this mean is that the p orbital is broken down into three different sub orbitals which is the Px , Py and Pz. Hence we can see that there are 6 electrons to enter into the P orbital too.

We can see that all the S orbitals have been completely filled with two electrons alike each. This is also the case for the P orbital as the 3 suborbitals take in 2 each to give a total of six

4 0
4 years ago
What is the specific heat of a 75.01 g piece of an unknown metal that exhibits a 45.2°C temperature change upon absorbing 1870 J
mariarad [96]

Answer:

Cp=0.552\frac{J}{g\°C}

Explanation:

Hello,

In this case, the formula to compute the required heat based on the mass, specific heat and change in temperature is widely known as:

Q=mCp\Delta T

In such a way, since we are asked to compute the specific heat, we solve for it as shown below:

Cp=\frac{Q}{m\Delta T}=\frac{1870J}{75.01g*45.2\°C} \\ \\Cp=0.552\frac{J}{g\°C}

Best regards.

8 0
3 years ago
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