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Natalka [10]
3 years ago
8

Why is a very small amount of cerium oxide nanoparticles need: A) the nanoparticles are elements

Chemistry
1 answer:
irina [24]3 years ago
7 0
Have a high suface area to volume ratio
You might be interested in
Which element has a larger atomic radius than sulfur? chlorine cadmium fluorine oxygen
zhuklara [117]

Answer:

  • <u>Cadmium has larger atomic radius than sulfur.</u>

Explanation:

Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.

Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:

  • Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.

  • Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.

Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:

  • Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.

  • Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.

So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:

  • O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.

Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.

8 0
3 years ago
How can you differentiate the Atomic Structure and the Bohr Model? In other words, how is the Atomic Structure different from th
Tatiana [17]

Answer:

Bohr thought that electrons orbited the nucleus in circular paths; whereas in the modern view atomic electron structure is more like 3D standing waves. Bohr built upon Rutherford's model of the atom. ... He believed that electrons moved around the nucleus in circular orbits with quantised potential and kinetic energies.

Explanation:

8 0
3 years ago
Determining nitrogen balance in the clinical setting can be very valuable. To determine a client’s state of nitrogen balance, a
nikdorinn [45]

Answer:

We also need the nitrogen intake

Explanation:

The nitrogen balance is the difference between the nitrogen intake and nitrogen excreted. If this calculation is positive, the body grow; if it is negative, the body is decreased.

6 0
3 years ago
how many oxygen atoms are there in six dinitrogen monoxide molecules? express your answer as an integer
stepan [7]

Answer:

6 oxygen atoms

Explanation:

Step 1: Given data

Number of dinitrogen monoxide molecules (N₂O): 6

Number of oxygen atoms (O): ?

Step 2: Calculate the appropriate ratio

The ratio of dinitrogen monoxide molecules to oxygen atoms is 1:1.

Step 3: Use the ratio to calculate the number of oxygen atoms

6 molecule N₂O × (1 atom O/1 molecule N₂O): 6 atom O

7 0
3 years ago
There are two naturally occurring isotopes of boron. 10 B has a mass of 10.0129 u. 11 B has a mass of 11.0093 u. Determine the a
Vanyuwa [196]
<h2>Natural Abundance for 10B is 19.60%</h2>

Explanation:

  • The natural isotopic abundance of 10B is 19.60%.
  • The natural isotopic abundance of 11B is 80.40%.
  • The isotopic masses of boron are 10.0129 u and 11.009 u respectively.

For calculation of abundance of both the isotopes -

Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.

Determining it as an equation -

10x + 11y= 10.8

x+y=1 (ratio)

10x + 10y = 10

By taking the denominator away from the numerator

we get;

y = 0.8

x + y = 1

∴ x = 0.2

To get percentages  we need to multiply it by 100

So, the calculated abundance is 80% for 11 B and 20% 10  B.

5 0
3 years ago
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