solution:
Weight of caffeine is W = 0.170 gm.
Volume of water is V= 10 ml
Volume of methylene chloride which extracted caffeine is v= 5ml
No of portions n=3
Distribution co-efficient= 4.6
Total amount of caffeine that can be unextracted is given by
![w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms](https://tex.z-dn.net/?f=w_%7Bn%7D%3Dw%5Ctimes%5B%5Cfrac%7Bk_%7BDx%7Dv%7D%7Bk_%7BDx%7Dv%2Bv%7D%5D%5En%5C%5C%3C%2Fp%3E%3Cp%3Ew_%7B3%7D%3D0.170%5B%5Cfrac%7B4.6%5Ctimes10%7D%7B%284.6%5Ctimes10%2B5%29%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B46%7D%7B46%2B5%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B46%7D%7B51%7D%5D%5E3%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5B%5Cfrac%7B97336%7D%7B132651%7D%5D%5C%5C%3C%2Fp%3E%3Cp%3E%3D0.170%5Ctimes0.734%3D0.125gms)
amount of caffeine un extracted is 0.125gms
amount of caffeine extracted=0.170-0.125
=0.045 gms
Answer:
Explanation:
Hello!
In this case, since the chemical reaction between copper and nitric acid is:
By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:
However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:
Best regards!
The heat of combustion for methanol is 727 kj/mol
<em><u>calculation</u></em>
calculate the moles of methanol (CH3OH)
moles = mass/molar mass
molar mass of methanol = 12 +( 1 x3) +16 + 1= 32 g /mol
moles is therefore= 64.0 g / 32 g/mol = 2 moles
Heat of combustion is therefore = 1454 Kj / 2 moles = 727 Kj/mol
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is
H = mcΔt
H - heat energy
m - mass of material
c - specific heat of the material
Δt - change in temperature
substituting the values in the equation
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
Coulomb's law mathematically is:
F = kQ₁Q₂/r²
we integrate this with respect to distance to obtain the expression for energy:
E = kQ₁Q₂/r; where k is the Coulomb's constant = 9 x 10⁹; Q are the charges, r is the seperation
Charge on proton = charge on electron = 1.6 x 10⁻¹⁹ C
E = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹) / (185 x 10⁻¹²)
E = 1.24 x 10⁻¹⁸ Joules per proton/electron pair
Number of pairs in one mole = 6.02 x 10²³
Energy = 6.02 x 10²³ x 1.24 x 10⁻¹⁸
= 746.5 kJ
