An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. A covalent bond involves a pair of electrons being shared between atoms.
The name of this alkane is with central carbons are bonded to c h 3 is 2-methylbutane.
<h3>
What is alkane?</h3>
Alkanes belong to the family of saturated hydrocarbons with carbon carbon single bond.
For the given alkane;
CH₃ H
CH₃ - C - C - CH₃
H H
Thus, the name of this alkane is with central carbons are bonded to c h 3 is 2-methylbutane.
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Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
Also,
For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.
Thus,
.................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
...................2
Given that:

To find,

Using 1 and 2 , we get:

<u>Size of the potassium ion = 1.33 Å</u>
Answer:
32, 30 and 41
Explanation:
The problem here is to find the number of:
Protons, neutrons and electrons in Ge²⁺
In this ion,
We must understand that for a net positive charge to remain on an atom, the number of protons must be greater than the number of electrons.
Ge is Germanium with atomic number of 32;
So the number of protons is 32
Since the atom has lost two electrons;
Number of electrons now is 32 - 2 = 30
Number of neutrons is 41 from the periodic table.
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M