Answer:
49.09 moles of gas are added to the container
Explanation:
Step 1: Data given
Initial volume = 3.10 L
Number of moles gas = 9.51 moles
The final volume = 19.1 L
The pressure, temperature remain constant
Step 2: Calculate number of moles gas
V1/n1 = V2/n2
⇒with V1 = the initial volume of the gas = 3.10 L
⇒with n1 = the initial number of moles = 9.51 moles
⇒with V2 = the increased volume = 19.1 L
⇒with n2 = the final number of moles gas
3.10L / 9.51 moles = 19.1 L / n2
n2 = 58.6 moles
The new number of moles is 58.6
Step 3: calculate the number of moles gas added
Δn = 58.6 - 9.51 = 49.09 moles
49.09 moles of gas are added to the container
Answer:
O-H bond
Explanation:
Let us work out the electronegativity difference between the elements in each bond in order to decide which of them is most polar.
For the C-O bond
2.55 - 2.2 =0.35
For the F-F bond
3.98 - 3.98 = 0
For the O-H bond
3.44 - 2.2 = 1.24
For the N-H bond
3.04 - 2.2 = 0.84
The O-H bond has the highest electronegativity difference, hence it is he most polar bond.
Answer:
It bonds with the added H+ or OH in solution.
Explanation:
Boyle’s Law illustrates the inverse relationship of volume and pressure. It follows the formula p1V1 = P2V2 , where P1V1 denotes initial pressure and volume and P2V2 denotes values of pressure and volume.
Now, let us work out for what is asked above.
a. if the pressure is doubled
50.0 p = V x 2p
V = 50.0 p / 2p
= 50.0 /2
= 25.0 m^3
b. if the pressure is cut in half
50.0 p = V x p/2
100 p = V x p
V = 100 m^3
c. if the pressure is tripled
50.0 p = V x 3p
V = 50.0 p / 3p
= 50.0 /3
=16.7 m^3
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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
