Answer:
D
Explanation:
it goes past the decimal three places
Answer:
<u>Optical purity = 76.9231 %</u>
<u>Specific rotation of mixture = - 97.6923 °</u>
Explanation:
The mass of the racemic mixture = 3 g
It means it contains R enantiomer = 1.5 g
S enantiomer = 1.5 g
Amount of Pure R = 10 g
Total R = 11.5 g
Total volume = 500 mL + 500 mL = 1000 mL = 1 L
[R] = 11.5 g/L
[S] = 1.5 g/L
Enantiomeric excess = 
= 
= 76.9231 %
<u>Optical purity = 76.9231 %</u>
Also,
Optical purity = 
Optical rotation of pure enantiomer = −127 °
<u>Specific rotation of mixture = - 97.6923 °</u>
Answer:
We need 1.1 grams of Mg
Explanation:
Step 1: Data given
Volume of water = 78 mL
Initial temperature = 29 °C
Final temperature = 78 °C
The standard heats of formation
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Step 2: The equation
The heat is produced by the following reaction:
Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)
Step 3: Calculate the mass of Mg needed
Using the standard heats of formation:
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)
−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg
(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required
(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg
We need 1.1 grams of Mg
Answer: 177g
Explanation:
Aw 12 = 6.02214076*10^23 atoms
mass = 12*88.70^23/6.022*10^23
