<u>We are given:</u>
Mass of water: 119 grams
We know that one mole of a gas occupies 22.4L of volume
<u>Number of moles of water:</u>
Number of moles = given mass / Molar mass
Number of moles = 119 / 18 [molar mass of water = 18 grams/mol]
Number of moles = 6.61 moles
<u>Volume occupied:</u>
Volume = number of moles * 22.4 L
Volume = 6.61 * 22.4
Volume = 148L
Volume (in mL) = 1.48 * 10⁻¹ mL
The energy required to separate a mole of an ionic solid into gaseous ions
Answer:
Explanation:
We have the reactions:
A: 
B: 
Our <u>target reaction</u> is:
We have 
as a reactive in the target reaction and is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.
A: 
Then if we add reactions A and B we can obtain the target reaction, so:
A:
B: 
For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.
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
