Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
Answer:
Volume of carbon dioxide is 428.23 L.
Explanation:
Below is the chemical reaction or chemical equation for the combustion of hydrocarbons such as undecane into carbon dioxide.
Here, undecane is in liquid form that reacts with gaseous oxygen (combustion) and produces carbon dioxide and water as a product in the gaseous form.
The molar mass of undecane = 
From the equation, it can be seen that 1 mole of undecane produces 11 moles of carbon dioxide. Therefore, 1.66 mol will produce 18.26 mol of carbon dioxide.
Now find the volume of 18.26 mol of carbon dioxide when the temperature is 13 degrees Celsius and pressure is 1 atm.
The hydrocarbon is used in excess.
<h3><u>Explanation</u>:</h3>
The bromination of an arene is not simple as bromination of an alkane. This is because the carbocation or free radicle formation in benzene is a very energy consuming process. This is why a lewis base like aluminium bromide or ferric bromide is used. The ferric bromide takes in the bromine radicle and forms the brominium cation which helps in the formation of electrophile. Now this electrophile brominium cation attacks the benzene ring and forms a temporary sp3 hybrid carbon intermediate. Then the hydrogen is taken by the FeBr4- forming HBr and regenerating the FeBr3 as well as Aromaticity of the arene species at the same time. Here hydrocarbon is used in excess just to prevent the chances of multiple substitution in the same arene molecule.
Answer:
Rb: [Kr] 5s
Step-by-step explanation:
Rb is element 37, the first element in Period 5.
It has one valence electron, so its valence electron configuration is 5s.
The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.
The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.
