Answer:1) Volume of 
required is 55.98 mL.
2) 0.62577 grams of 
is produced.
Explanation:
1) Molarity of 
Volume of 
Molarity of 
Volume of 
According to reaction, 1 mole of 
reacts with 3 mole of , then, 0.0041985 moles of will react with:
moles of that is 0.0125955 moles.
Volume of required is 55.98 mL.
2)
Number of moles of 
According to reaction, 3 moles of gives 1 mole of , then 0.004485 moles of will give:
moles of that is 0.001495 moles.
Mass of =
Moles of × Molar Mass of
= 0.001495 moles × 418.58 g/mol = 0.62577 g
0.62577 grams of is produced.
False.
Hope this helps, Good luck on the assignment.
Since the addition of the H2O in the last step of hydroboration is anti-Markovnikov, the starting material is 1-pentyne.
The addition of H2 to C5H8 yields an alkene when a Lindlar catalyst is used. Recall that the Lindlar catalysts poisons the process so that the addition do not go on to produce an alkane.
When hydroboration is carried out on the alkene, we are told that a primary alcohol was obtained. We must note that in the last step of hydroboration, water is added in an anti- Markovnikov manner to yield the primary alcohol. Hence, the starting material must be 1-pentyne as shown in the image attached.
Learn more: brainly.com/question/2510654
Answer:
2.79 °C/m
Explanation:
When a nonvolatile solute is dissolved in a pure solvent, the boiling point of the solvent increases. This property is called ebullioscopy. The temperature change (ΔT) can be calculated by:
ΔT = Kb*W*i
Where Kb is the ebullioscopy constant for the solvent, W is the molality and i is the van't Hoff factor.
W = m1/(M1*m2)
Where m1 is the mass of the solute (in g), M1 is the molar mass of the solute, and m2 is the mass of the solvent (in kg).
The van't Hoff factor represents the dissociation of the elements. For an organic molecule, we can approximate i = 1. Thus:
m1 = 2.00 g
M1 = 147 g/mol
m2 = 0.0225 kg
W = 2/(147*0.0225)
W = 0.6047 mol/kg
(82.39 - 80.70) = Kb*0.6047*1
0.6047Kb = 1.69
Kb = 2.79 °C/m
<em>Maybe</em>
<em>Mg+</em><em>Fe2O3</em><em>→</em><em>F</em><em>e</em><em>+</em><em>MgO</em>
<em>hope</em><em> it</em><em> helps</em>
