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
A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two.
Answer :
AgI should precipitate first.
The concentration of 
when CuI just begins to precipitate is, 
Percent of remains is, 0.0076 %
Explanation :
for CuI is 
for AgI is 
As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.
Now we have to calculate the concentration of iodide ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
![1\times 10^{-12}=0.0079\times [I^-]](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-12%7D%3D0.0079%5Ctimes%20%5BI%5E-%5D)
![[I^-]=1.25\times 10^{-10}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D1.25%5Ctimes%2010%5E%7B-10%7DM)
Now we have to calculate the concentration of silver ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ag^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BI%5E-%5D)
![8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M](https://tex.z-dn.net/?f=8.3%5Ctimes%2010%5E%7B-17%7D%3D%5BAg%5E%2B%5D%5Ctimes%201.25%5Ctimes%2010%5E%7B-10%7DM)
![[Ag^+]=6.64\times 10^{-7}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D6.64%5Ctimes%2010%5E%7B-7%7DM)
Now we have to calculate the percent of remains in solution at this point.
Percent of remains = 
Percent of remains = 0.0076 %
Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.
<h3>Answer:</h3>
a) Moles of Caffeine = 1.0 × 10⁻⁴ mol
b) Moles of Ethanol = 4.5 × 10⁻³ mol
<h3>Solution:</h3>
Data Given:
Mass of Caffeine = 20 mg = 0.02 g
M.Mass of Caffeine = 194.19 g.mol⁻¹
Molecules of Ethanol = 2.72 × 10²¹
Calculate Moles of Caffeine as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 0.02 g ÷ 194.19 g.mol⁻¹
Moles = 1.0 × 10⁻⁴ mol
Calculate Moles of Ethanol as,
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Molecules ÷ 6.022 × 10²³
Putting values,
Number of Moles = 2.72 × 10²¹ Molecules ÷ 6.022 × 10²³
Number of Moles = 4.5 × 10⁻³ Moles
