In Williamson ether synthesis, the reaction begins when the hydrogen from the alcohol's hydroxyl group contacts the halogen ion. Since the iodide ion is larger than the chloride ion, there is a larger chance of collision between it and the hydrogen atom, so butyl iodide is preferred over butyl chloride.
Also, Williamson synthesis is carried out at relatively high temperatures, around 50ºC-100º C. Ethyl iodide has a boiling point of 72.2ºC, so it will be in the liquid phase. Ethyl chloride has a boiling point of 12.3ºC so it will be in the gaseous phase, which is undesirable for this.
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
The density of gasoline is 0.7 g/cm3, and the density of water is 1 g/cm3. Thus the mass of the gasoline is 55*0.7 = 38.5g and the mass of the water is 60g.
Combining the 55 cm3 and 60 cm3 of substances with the aforementioned masses yields a volume of 55+60 = 115cm3 and a mass of 38.5+60 = 98.5g. The density is therefore 98.5/115 = 0.86 g/m3.
Answer:
The third line
Explanation:
The first one us wrong because they're not all found inside,electron is outside.
The second one is wrong as I explained previously.
Lastly the fourth line is wrong because protons are found in the nucleus.
To determine the amount of oxygen that is present in the compound, we have to assume that the given compound contains carbon, hydrogen and oxygen only or else we will not be able to determine the answer. We need to convert the moles of the elements given to units of grams by using the atomic mass of these elements. Then, from the total amount of the compound we subtract the masses of the elements. We do as follows:
mass
0.117 mol C ( 12.01 g / 1 mol ) = 1.41 g
0.233 mol H ( 1.01 g / 1 mol ) = 0.24 g
Mass O = 3.50 g - 1.41 g - 0.24 g = 1.85 g O
Moles O = 1.85 g O ( 1 mol / 16 g ) = 0.116 moles O
