Assume it is 1 litre and weighs 1kg.
2 percent of 1 kg is 20g.
20g divided by molar mass of NaOH.
20g divide by 40 = 0.5 mole
0.5 mole in a litre would be 0.5M
That is the answer: 0.5M
The mixture contains 62 % one isomer and 38 % the enantiomer.
Let’s say that the mixture contains 62 % of the (<em>R</em>)-isomer.
Then % (<em>S</em>) = 100 % -62 % = 38 %
ee = % (<em>R</em>) - % (<em>S</em>) = 62 % -38 % = 24 %
The answer you are looking for is A. If you need me to show you how I got the answer let me know. :)
The question does not provide the equation
Answer:-
72.89 grams
Explanation:-
The balanced chemical equation for this reaction is
CuSO4 + Fe --> FeSO4 + Cu
Molecular weight of CuSO4 = 63.55 x 1 + 32 x 1 + 16 x 4
= 159.55 gram
Atomic weight of Cu = 63.55 gram.
According to the balanced chemical equation
1 CuSO4 gives 1 Cu
∴159.55 gram of CuSO4 would give 63.55 gram of Cu.
183 gram of CuSO4 would give 63.55 x 183 / 159.55
= 72.89 grams of Cu
Given
Mass of NO - 824 g
Molar mass of NO - 30.01g/mol
No of moles of NO = Given mass/Molar mass
No of moles of NO = 824/30.01= 27.45 mole
Hence 27.5 moles of NO are formed!
