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3 years ago

Which of the following atoms has the greatest atomic radius? Oxygen (o) fluorine (F) Chlorine (CL) Carbon (C)

Chemistry

1 answer:

Andrei [34K]3 years ago

5 0

Periodic Trend:

The Atomic radius of atoms generally decreases from left to right across a period

Group Trend:

The atomic radius of atoms generally increases from top to bottom within a group. As atomic number increases down a group, there is a increase in the positive nuclear charge, however the co-occurring increase in the number of orbitals wins out, increasing the atomic radius down a group in the periodic table

Answer :

The Atom with the greatest atomic radius is chlorine. Fluorine can be ruled out because it is in the same period as oxygen and further to the right down the period. Chlorine has the largest atomic size because it is farthest down the group of any of the above elements listed.

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The density of gas particles in a section of Earth’s atmosphere decreases. Which of the following is the most likely explanation

Bad White [126]

This is due to the fact that the air pressure in that certain section of Earth’s atmosphere decreased. As density of gas particles decreases as air pressure decreases. Therefore, density of gas particles and air pressure have a direct relationship. An increase in air pressure would then effect to an increase in gas particles.

6 0

3 years ago

11. What is the specific heat of a substance with a mass of 25.5 g that requires 412 J

Romashka-Z-Leto [24]

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of

1 g

of a given substance by

∘

In your case, aluminium is said to have a specific heat of

0.90

∘

So, what does that tell you?

In order to increase the temperature of

1 g

of aluminium by

∘

, you need to provide it with

0.90 J

of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by

∘

. So if you wanted to increase the temperature of

10.0 g

of aluminium by

∘

, you'd have to provide it with

1 gram



0.90 J

1 gram



0.90 J

...

1 gram



0.90 J



10 times

0.90 J

However, you don't want to increase the temperature of the sample by

∘

, you want to increase it by

∘

−

∘

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

∘



0.90 J

∘



0.90 J

...

∘



0.90 J



33 times

0.90 J

Therefore, the total amount of heat needed to increase the temperature of

10.0 g

of aluminium by

∘

will be

10.0

⋅

0.90

∘

⋅

∘

297 J

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

⋅

, where

- the amount of heat added / removed

- the mass of the substance

- the specific heat of the substance

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

6 0

4 years ago

A biochemist carefully measures the molarity of magnesium ion in of cell growth medium to be . Unfortunately, a careless graduat

denis23 [38]

Answer:

760 uM

Explanation:

A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.

The problem here is that the amount of magnesium ion remains the same irrespective of the volume.

Amount of magnesium in the growth medium = molarity x volume

= 97 x $10^{-6}$ x 47 x $10^{-3}$ = 4.559 x $10^{-6}$

Then, the volume reduced to 6.0 mL, the new molarity becomes;

molarity = mole/volume

= 4.559 x $10^{-6}$ /6 x $10^{-3}$ = 7.598333 x $10^{-4}$ M = 759.83333 uM

To the correct number of significant digits = 760 uM

6 0

3 years ago

What is the binding energy of a mole of nuclei with a mass defect of 0.00084 kg/mol?

lana [24]

Answer:

7.56 x 10^13 J/mol

Explanation:

Binding energy = (Mass defect) × (velocity of light)² = (84 × 10^-5) × (3 × 10^8)^2 = 756 × 10^11 J = 7.56 × 10^13 J/mol

Hope this helps -Paige

7 0

3 years ago

T or F: Lone pairs around the oxygen atom of a water molecule play no role in determining its molecular geometry?

Mama L [17]

Answer:

Explanation:

They play a very important part. The geometry is not a straight line. It is an angle over 90 which means that the molecule has the same general shape as a boomerang. The two hydrogens and the 2 lone electron pairs try to get away as far as possible from each other. The actual shape results in a tetrahedron shape. But the two hydrogens and 1 oxygen actually look like the aforementioned boomerang.

5 0

2 years ago