Complete Question
A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane
The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?
Answer:
The mass of caffeine extracted is 
Explanation:
From the question above we are told that
The K value is 
The mass of the caffeine is 
The volume of water is 
The volume of caffeine is 
The number of times the extraction was done is n = 3
Generally the mass of caffeine that will be extracted is
![P = m * [\frac{V}{K * v_c + V} ]^3](https://tex.z-dn.net/?f=P%20%3D%20%20m%20%20%2A%20%20%5B%5Cfrac%7BV%7D%7BK%20%2A%20%20v_c%20%2B%20V%7D%20%5D%5E3)
substituting values
![P = 40 * [\frac{2}{4.6 * 2 + 2} ]^3](https://tex.z-dn.net/?f=P%20%3D%20%2040%20%20%20%2A%20%20%5B%5Cfrac%7B2%7D%7B4.6%20%2A%20%202%20%2B%202%7D%20%5D%5E3)
Answer:
See explanation
Explanation:
The principle of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another. Hence, chemical energy in a battery can be converted to electrical energy.
Usually, the conversion of energy from one form to another is not 100% efficient according to the second law of thermodynamics. Some energy is wasted in the process, sometimes as heat.
Hence, in an ideal situation where no heat energy is produced; all the chemical energy is converted to electrical energy (100% energy conversion). There will be no energy loss if no heat is produced.
Answer:
25
Explanation:
explain no need it is perfect
C. 2 hydrogen (H) atoms because in bonding with them sulfur will get a full valence shell and hydrogen will have a full valence shell.
Answer:
A) 0.1225 M
B) 100.4 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA(aq) + NaOH(aq) ⇒ NaA(aq) + H₂O(l)
Step 2: Calculate the reacting moles of NaOH
17.73 mL of 0.1036 M NaOH react. The reacting moles are:
0.01773 L × 0.1036 mol/L = 1.837 × 10⁻³ mol
Step 3: Calculate the reacting moles of HA
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA are 1/1 × 1.837 × 10⁻³ mol = 1.837 × 10⁻³ mol.
Step 4: Calculate the molar concentration of HA
1.837 × 10⁻³ moles of HA are in a 15.00 mL volume. The molar concentration is:
M = 1.837 × 10⁻³ mol / 0.01500 L = 0.1225 M
Step 5: Calculate the molar mass of HA
1.837 × 10⁻³ moles of HA weigh 0.1845 g. The molar mass of HA is:
0.1845 g / 1.837 × 10⁻³ mol = 100.4 g/mol
