was is your question on it and what grade level are you?
q = mCΔT
The correct specific heat capacity of water is <em>4.187 kJ/(kg.K)</em>.
ΔT = q/mC = 87 kJ/[648.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2713 kJ/K) = 0.032 K
Tf = Ti + ΔT = 298 K + 0.032 K = 298.032 K
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) = ![\frac{C_1}{C_2}](https://tex.z-dn.net/?f=%5Cfrac%7BC_1%7D%7BC_2%7D)
where;
solubility of compound in the organic solvent and
= solubility in aqueous water.
So; we can represent our data as:
÷ ![(\frac{B_{(mg)}}{100mL} )](https://tex.z-dn.net/?f=%28%5Cfrac%7BB_%7B%28mg%29%7D%7D%7B100mL%7D%20%29)
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=
÷ ![(\frac{B_{(mg)}}{100mL})](https://tex.z-dn.net/?f=%28%5Cfrac%7BB_%7B%28mg%29%7D%7D%7B100mL%7D%29)
4.6 = ![\frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}](https://tex.z-dn.net/?f=%5Cfrac%7B%28%5Cfrac%7B0.6-B%28mg%29%7D%7B60mL%7D%20%29%7D%7B%28%5Cfrac%7BB_%7B%28mg%29%7D%7D%7B100mL%7D%29%7D)
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 = ![\frac{0.6}{3.76}](https://tex.z-dn.net/?f=%5Cfrac%7B0.6%7D%7B3.76%7D)
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.
Li2S + 2 HNO3 --> 2 LiNO3 + H2S
Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S
Li S + H2 N2 O5 -> Li N2 O5 + H2 S
Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2
Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2