I really don’t know but I think you should be more specific
Answer:
Look at the picture.
Explanation:
(2S,3S)-2-Bromo-3-phenylbutane will undergo E2 reaction and form trans product of elimination due to its thermodynamic stability.
Answer:
8.37 grams
Explanation:
The balanced chemical equation is:
C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)
Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.
From the ideal gas law we can determine the number of moles that the 2.25 L represent.
From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.
PV = nRT ⇒ n = PV/RT
n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂
Moles glucose required:
0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆
The molar mass of glucose is 180.16 g/mol, then the mass required is
0.046 mol x 180.16 g/mol = 8.37 g
Each correspond to a principal energy level
Answer:
Explanation:
A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 Go to calculating final temperature when mixing two samples of water ... Problem #1: A 610. g piece of copper tubing is heated to 95.3 °C and placed in an ... The two rings are heated to 65.4 °C and dropped into 12.4 mL of water at 22.3 °C. ... Problem #4: A 5.00 g sample of aluminum (specific heat capacity = 0.89 J g¯1
