Answer : The % of (+) limonene isomer = 79%
The % of (-) limonene isomer = 0%
The % of enantiomeric excess = 58%
Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.
Given,
% of pure limonene enantiomer = The % of (+) limonene isomer = 79%
Therefore, The % of (-) limonene isomer = 0%
Formula used :

Where, ee → enantiomeric excess
Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


= 58%
Answer: 2.7moles
Explanation:
6Cs + Fe2(CO3)3 —> 3Cs2CO3 + 2Fe
From the equation,
6moles of Cs produced 3 moles of Cs2CO3.
Therefore, 5.34 moles of Cs will produce = (5.4x3)/6 = 2.7moles of Cs2CO3
Answer:
Explanation:
Percent composition is percentage by the mass of element present in the compound.
The formula for chromium(III) nitrate is 
Molar mass of chromium(III) nitrate = 238.011 g/mol
1 mole of chromium(III) nitrate contains 9 moles of oxygen
Molar mass of oxygen = 16 g/mol
So, Mass= Molar mass*Moles = 16*9 g = 144 g
<span>Answer:
Zn(2+) + 2e(-) -------> Zn
1 mole of Zn is deposited by 2F of electricity ...
so 48.9 mole of Zn will be deposited by 48.9 X 2F = 97.8 F of electricity...
as 1F = 96500 C
so 97.8 F = 97.8 X 96500 = 9437700 C of electricity...</span>