Answer:
See explanation below
Explanation:
First, we need to understand that the monochlorination of an alkane like this one, involves substitution of one of the atoms of hydrogen of the molecule for an atom of chlorine.
This reaction takes place when the alkane reacts with Cl₂ in presence of light or heat.
When this happens, the first step involves the breaking of the double bond of the chlorine to form the ion Cl⁻.
The next step involves the substraction of the hydrogen of the molecule by the Chlorine. This will leave the alkane with a lone pair available for reaction.
The third step, the alkane with the lone pair of electron substract a chlorine for the beggining and form the mono chlorinated product.
The final step involves forming the remaining products with the remaining reagents there.
In the picture attached you have the mechanism and product for this reaction:
Answer: Electrons in bonding orbitals stabilize the molecule because they are between the nuclei. They also have lower energies because they are closer to the nuclei. Antibonding orbitals place less electron density between the nuclei. The nuclear repulsions are greater, so the energy of the molecule increases.
Explanation:
Answer:
Explanation:
1. Solubility of CaF_2
(a) Molar solubility
CaF₂ ⇌ Ca²⁺ + 2F⁻
![K_{\text{sp }} = \text{[Ca$^{2+}$]}\text{[F$^{-}$]}^{2}= 4.0 \times 10^{-8}\\s(2s)^{2}=4.0 \times 10^{-8}\\4s^{3} = 4.0 \times 10^{-8}\\s^{3} = 1.0 \times 10^{-8}\\s =2.2 \times 10^{-3}\text{ mol/L}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bsp%20%7D%7D%20%3D%20%5Ctext%7B%5BCa%24%5E%7B2%2B%7D%24%5D%7D%5Ctext%7B%5BF%24%5E%7B-%7D%24%5D%7D%5E%7B2%7D%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%282s%29%5E%7B2%7D%3D4.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5C4s%5E%7B3%7D%20%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%5E%7B3%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-8%7D%5C%5Cs%20%3D2.2%20%5Ctimes%2010%5E%7B-3%7D%5Ctext%7B%20mol%2FL%7D)
(b) Mass solubility
2. pH
pH = -log [H⁺] = -log(3.0 × 10⁻⁴) = 3.52
3. Oxidizing and reducing agents
Zn + Cl₂ ⟶ ZnCl₂
The oxidation number of Cl has decreased from 0 to -1.
Cl has been reduced, so Cl is the oxidizing agent.
4. Oxidation numbers
(a) Al₂O₃
1O = -2; 3O = -6; 2Al = +6; 1Al = +3
(b) XeF₄
1F = -1; 4F = -4; 1 Xe = +4
(c) K₂Cr₂O₇
1K = +1; 2K = +2; 1O = -2; 7O = -14
+2 - 14 = -12
2Cr = + 12; 1 Cr = +6
Answer:
V = 16.81 L
Explanation:
Given data:
Number of moles = 0.75 mol
Temperature = standard = 273 K
Pressure = 1 atm
Volume of balloon = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will rearrange the formula.
V = nRT/P
V = 0.75 mol ×0.0821 atm.L/mol.K ×273 K / 1 atm
V = 16.81 atm.L/ 1 atm
V = 16.81 L
Answer:
1281.25 Moles of glucose
Explanation:
Multiply 1.25x1025= 1281.25
