Answer:
74.0 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
At the equivalence point, 33.83 mL of 0.115 M NaOH react.
0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol
Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.
Step 4: Calculate the molar mass of the acid
3.89 × 10⁻³ moles of HA have a mass of 0.288 g.
M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol
6.022x10^23 is Avogadro’s number. Use this whenever you work with Stoichiometry involving Atoms, formula units, or molecules. 1 mol of anything is always Avogadro’s number.
Multiply everything on the top= 6.93 x 10^23
Divide by everything on the bottom = 6.93 x 10^23
Answer: 6.93 x 10^23 atoms Cu.
Answer:
I wrote this answer to get points lol
Explanation:
Wind abrades rock by sandblasting, this is the process in which wind causes the
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.