Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
Answer:
1.95
Explanation:
Just did the test and it was correct!
Answer is: the absolute pressure of the air in the balloon is 1.015 atm (102.84 kPa).
n = 0.250 mol; amount of substance.
V = 6.23 L; volume of the balloon.
T = 35°C = 308.15 K; temperature.
R = 0.08206 L·atm/mol·K, universal gas constant.
Ideal gas law: p·V = n·R·T.
p = n·R·T / V.
p = 0.250 mol · 0.08206 L·atm/mol·K · 308.15 K / 6.23 L.
p = 1.015 atm; presure of the air.
Bromine (Br) is the only element listed that is a diatomic element. I believe that Bromine is going to be your answer.
Answer:
0.256 L
Explanation:
We should use the following formula:
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.82 M NaOCl
volume (1) = ?
concentration (2) = 0.21 M NaOCl
volume (2) = 1 L
volume (1) = [concentration (2) × volume (2)] / concentration (1)
volume (1) = [0.21 / 1] / 0.82 = 0.256 L
