Answer:
B. Similar fossils were found in different continents.
Answer:
![[I_2]=[Br]=0.31M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%5BBr%5D%3D0.31M)
Explanation:
Hello there!
In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:
Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:
![K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%5C%5C%5C%5C1.2x10%5E2%3D%5Cfrac%7B%282x%29%5E2%7D%7B%282.0-x%29%5E2%7D)
Thus, we solve for x as show below:
Therefore, the concentrations of both bromine and iodine are:
![[I_2]=[Br]=2.0M-1.69M=0.31M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%5BBr%5D%3D2.0M-1.69M%3D0.31M)
Regards!
Answer:
The answer to your question is: kc = 6.48
Explanation:
Data
Given Molecular weight
CaO = 44.6 g 56 g
CO₂ = 26 g 44 g
CaCO₃ = 42.3 g 100 g
Find moles
CaO 56 g ---------------- 1 mol
44.6 g -------------- x
x = (44.6 x 1) / 56 = 0.8 mol
CO₂ 44 g ----------------- 1 mol
26 g ---------------- x
x = (26 x 1 ) / 44 = 0.6 moles
CaCO₃ 100 g --------------- 1 mol
42.3g -------------- x
x = (42.3 x 1) / 100 = 0.423 moles
Concentrations
CaO = 0.8 / 6.5 = 0.12 M
CO₂ = 0.6 / 6.5 = 0.09 M
CaCO₃ = 0.423 / 6.5 = 0.07 M
Equilibrium constant = ![\frac{[products]}{[reactants]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bproducts%5D%7D%7B%5Breactants%5D%7D)
Kc = [0.07] / [[0.12][0.09]
Kc = 0.07 / 0.0108
kc = 6.48
548.6 it would be the same thing
Clockwise from the carbon connected to three H atoms:
- C: sp³
- N: sp³
- C: sp³
- C: sp²
- N: sp².
<h3>Explanation</h3>
Start by finding the number of electron domains on each C and N atom. Why the number of electron domains? The number of electron domains of an atom indicates it hybridization. For atoms in period two (which includes both C and N):
- An atom with four electron domains is sp³ hybridized.
- An atom with three electron domains is sp² hybridized.
- An atom with two electron domains is sp hybridized.
How many electron domains on each of the atoms?
For each atom:
- Each atom that the atom in question is connected to (via covalent bonds, for sure) counts towards one electron domain. This rule shall hold for bonds of all orders. (i.e., No matter if the the two atoms are connected via a Single bond, a double bond, or a triple bond.) In other words, each C-C or C-N single bond counts towards one electron domain. Each C=N double bond also counts towards one electron domain.
- Each lone pair on the atom in question counts towards one electron domain. Keep in mind that there are two electrons in one lone pair. (Hence the name "pair".)
For example:
- The carbon atom at the bottom of the graph is connected to four other atoms- three Hs and one N. There's no lone pair on that atom. That C atom contains four electron domains, which implies that the atom is sp³ hybridized.
- The nitrogen atom near the right end of the molecule is connected to two other atoms- one C and one H. There's one lone pair on that molecule. 2 + 1 = 3. That N atom contains three electron domains, which implies that the atom is sp² hybridized.
Try to figure out the number of electron domains on the rest of the atoms. Then determine their hybridization. In conclusion, clockwise from the carbon connected to three H atoms:
- C: sp³
- N: sp³
- C: sp³
- C: sp²
- N: sp².
