Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
Answer:
<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>
<em>b) 8.69 x 10^7 lb/ft^3</em>
<em></em>
Explanation:
mass of the star m = 2.0 x 10^36 kg
radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m
The density of substance ρ = mass/volume
The volume of the star = volume of a sphere = 
==> V = 
= 1.437 x 10^27 m^3
density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3
in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>
in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>
Answer:
Divide the mass of your anhydrous (heated) salt sample by the molar mass of the anhydrous compound to get the number of moles of compound present. In our example, 16 grams / 160 grams per mole = 0.1 moles. Divide the mass of water lost when you heated the salt by the molar mass of water, roughly 18 grams per mole.In order to determine the formula of the hydrate, [Anhydrous Solid⋅xH2O], the number of moles of water per mole of anhydrous solid (x) will be calculated by dividing the number of moles of water by the number of moles of the anhydrous solid (Equation 2.12. 6).
Answer:
0,508g of H₂O₂
Explanation:
For the reaction:
2KMnO₄(aq) + H₂O₂(aq) + 3H₂SO₄(aq) → 3O₂(g) + 2MnSO₄(aq) + K₂SO₄(aq) + 4H₂O(l)
2 moles of KMnO₄ react with 1 mol of H₂O₂.
In the titration, moles of KMnO₄ required were:
1,68M×0,0178L = 0,0299 moles of KMnO₄. Moles of H₂O₂ are:
0,0299 moles of KMnO₄×
= 0,01495 moles of H₂O₂. As molar mass of H₂O₂ is 34,01g/mol, mass of H₂O₂ was dissolved is:
0,01495 moles of H₂O₂×
= <em>0,508g of H₂O₂</em>
