Answer:
2-methoxy-2-methylpropane
Explanation:
The first step for this reaction is the carbocation formation. In this step, a tertiary carbocation is formed. Also, we will have a good leaving group so bromide will be formed. Then the methanol acts as a nucleophile and attacks the carbocation. Next, a positive charge is generated upon the oxygen, this charge can be removed when the hydrogen leaves the molecule as
. (See figure)
First convert grams to moles:
70.0g *(mole/98.079) = 0.7137mole
Remember that molarity is moles per liter:
0.7137mole *(1/280mL) *(1000mL/L) = 25.5M
Answer: The average atomic mass of X is 16.53
Explanation:
Mass of isotope X-19 = 2.282
% abundance of isotope X-19 = 12.01% =
Mass of isotope X-21 = 18.48
% abundance of isotope X-21 = (100-12.01)% =
Formula used for average atomic mass of an element :
Therefore, the average atomic mass of X is 16.53
62 g of water are vaporized and the energy required is 2.4 kJ/g
So 62g x 2.4 kJ/g = 148.8 kJ or 148,800 Joules
Q = mCΔT
Q is energy in joules, m is mass of water, C is the specific heat, delta T is change in temp
148,800 = m(4.18)(41 - 19) = 1618g or 1.6 kg of water
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane