Answer:
324.18 g/mol
Explanation:
Let the molecular mass of the antimalarial drug, Quinine is x g/mol
According to question,
Nitrogen present in the drug is 8.63% of x
So, mass of nitrogen = 
Also, according to the question,
2 atoms are present in 1 molecule of the drug.
Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)
So, mass of nitrogen = 14.01×2 = 28.02
These 2 must be equal so,

solving for x, we get:
<u>x = 324.18 g/mol</u>
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
First you need to find the amount of mass of Na2CO3 in one moles
(Use periodic chart)
Na= 22.99 x 2 = 45.98
C = 12.01
O = 16.00 x 4 = 64.00
Add the molar masses together to get 121.99
To find how many grams are in 4 moles, times 121.99 by 4
This gives you 487.96
But the questions asks for the answer to be in kilograms nor grams, to change into kilograms divide by 1000
This gets you the answer: 0.49 kg
Your answer is 3.80 moles
Answer:
The reducing agent is Zn.
Explanation:
Let's consider the reaction between zinc and hydrochloric acid.
Zn(s) + 2 HCl(aq) ⇄ ZnCl₂(aq) + H₂(g)
This is a redox reaction, which can be divided in 2 half-reactions: reduction and oxidation.
In the reduction, H⁺ gains electrons and it is considered the oxidizing agent.
2H⁺ + 2 e⁻ ⇒ H₂
In the oxidation, Zn loses electrons and it is considered the reducing agent.
Zn ⇒ Zn²⁺ + 2 e⁻