Answer:
See explanation and image attached
Explanation:
The Gilman reagent is a lithium and copper (diorganocopper) reagent with a general formula R2CuLi. R is an alkyl or aryl group.
They are useful in the synthesis of alkanes because they react with organic halides to replace the halide group with an R group.
In this particular instance, we intend to synthesize propylcyclohexane. The structure of the lithium diorganocopper (Gilman) reagent required is shown in the image attached to this answer.
Answer:
Each row is called a period. Each column is called a group or family.
Explanation:
A. Because sodium is a metal and metals are good conductors
Answer:
3.0 moles Al₂O₃
Explanation:
We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.
4 Al + 3 O₂ -----> 2 Al₂O₃
6.0 moles Al 2 moles Al₂O₃
---------------------- x ------------------------- = 3.0 moles Al₂O₃
4 moles Al
4.0 moles O₂ 2 moles Al₂O₃
---------------------- x ------------------------- = 2.7 moles Al₂O₃
3 moles O₂
As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.
However, since this problem is directly addressing how much Al₂O₃ is produced from Al, the answer you most likely are looking for is 3.0 moles Al₂O₃.
Answer:
13.20
Explanation:
Step 1: Calculate the moles of Ba(OH)₂
The molar mass of Ba(OH)₂ is 171.34 g/mol.
0.797 g × 1 mol/171.34 g = 4.65 × 10⁻³ mol
Step 2: Calculate the molar concentration of Ba(OH)₂
Molarity is equal to the moles of solute divided by the liters of solution.
[Ba(OH)₂] = 4.65 × 10⁻³ mol/60 × 10⁻³ L = 0.078 M
Step 3: Calculate [OH⁻]
Ba(OH)₂ is a strong base according to the following equation.
Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻
The concentration of OH⁻ is 2/1 × 0.078 M = 0.16 M
Step 4: Calculate the pOH
pOH = -log OH⁻ = -log 0.16 = 0.80
Step 5: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - 0.80 = 13.20
