Answer:
159 mg caffeine is being extracted in 60 mL dichloromethane
Explanation:
Given that:
mass of caffeine in 100 mL of water = 600 mg
Volume of the water = 100 mL
Partition co-efficient (K) = 4.6
mass of caffeine extracted = ??? (unknown)
The portion of the DCM = 60 mL
Partial co-efficient (K) = 
where;
solubility of compound in the organic solvent and
= solubility in aqueous water.
So; we can represent our data as:
÷ 
Since one part of the portion is A and the other part is B
A+B = 60 mL
A+B = 0.60
A= 0.60 - B
4.6=
÷ 
4.6 = 
4.6 ×
=
4.6 B
= 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B = 
B = 0.159 g
B = 159 mg
∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:

Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL = 
Thus, the fuel density of pentane is 
Answer:
A. 4.
Explanation:
<em>no. of moles (n) = mass/molar mass.</em>
mass = 64.0 g, molar mass = 16.0 g/mol.
∴ no. of moles (n) = mass/molar mass = (64.0 g)/(16.0 g/mol) = 4 mol.
Here is a universal law to balance chemical equations :
All chemical equations must be balanced
because of the law of conservation of mass.
It states that "matter
cannot be created or destroyed."
So, the number of atoms that you
start with at the beginning of the reaction must equal the number
of atoms that you end up with.
keeping this law in our mind,lets balance the<span> equation for the reaction of benzene and hydrogen to form cyclohexane.
</span><span>C 6 H 6 + H 2 → C 6 H 12.
Here is the balance chemical equation.
</span><span>2 C 6 H 6 + 6 H 2 → 2 C 6 H 12.</span>