Answer :
There is the commercial-grade, which is 70% strength in water, and it's pretty nasty stuff. It'll chew through your lab coat and give you burns you'll regret, as you'd expect from something that's rather stronger than nitric or sulfuric acid.
But it has other properties. The perchlorate anion is in a high oxidation state, and what goes up, must come down. A rapid drop in oxidation state, as chemists know, is often accompanied by loud noises and flying debris, particularly when the products formed are gaseous and have that pesky urge to expand. If you take the acid up to water-free concentrations, which is most highly not recommended, you'll probably want to wear chain mail, because it's tricky stuff. You can even go further and distill out the perchloric anhydride (dichlorine heptoxide) if you have no sense whatsoever. It's a liquid with a boiling point of around 80 C, and I'd like to shake the hand of whoever determined that property, assuming he has one left.
Answer:
b. primitive cubic < body-centered cubic < face-centered cubic
Explanation:
The coordination number is defined as <em>the number of atoms (or ions) surrounding an atom (or ion) in a crystal lattice</em>. Its value gives us a measure of how tightly the spheres are packed together. The larger the coordination number, the closer the spheres are to each other.
- In the <u>primitive cubic</u>, each sphere is in contact with 6 spheres, so its <u>coordination number is 6</u>.
- In the <u>body-centered cubic</u>, each sphere is in contact with 8 spheres, so its <u>coordination number is 12</u>.
- In the <u>face-centered cubic</u>, each sphere is in contact with 12 spheres, so its <u>coordination number is 12</u>.
Therefore, the increasing order in density is the primitive cubic first, then the body-centered cubic, and finally the face-centered cubic.
Answer:
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l) + 6 e⁻
Explanation:
In order to balance a half-reaction we use the ion-electron method.
Step 1: Write the half-reaction
Cr³⁺(aq) = Cr₂O₇²⁻(aq)
Step 2: Perform the mass balance, adding H₂O(l) and OH⁻(aq) where appropriate
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l)
Step 3: Perform the electric balance, adding electrons where appropriate
14 OH⁻(aq) + 2 Cr³⁺(aq) = Cr₂O₇²⁻(aq) + 7 H₂O(l) + 6 e⁻
Explanation: Outer shell electronic configuration for Group 18 is :
( have 8 electrons in the outer shell)
The electronic configurations for other groups are:
( have 1 electron in the outer shell}
( have 2 electrons in the outer shell)
( have 5 electrons in the outer shell)
( have 6 electrons in the outer shell)
( have 7 electrons in the outer shell)
Comparing the electron configurations of group 18 to other groups.
Group 18 has fully filled outer shell, that is 8 electrons in the outer shell, while others are partially filled.
<span>Two characteristics used to classify igneous rocks are texture and
</span>Mineral Grains
