Explanation:
high energy to Low energy
=the electron gains energy (K.E)
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
Answer:
3.18 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 0.985 atm
- Initial volume (V₁): 3.65 L
- Final pressure (P₂): 861.0 mmHg
Step 2: Convert P₁ to mmHg
We will use the conversion factor 1 atm = 760 mmHg.
0.985 atm × 760 mmHg/1 atm = 749 mmHg
Step 3: Calculate the final volume of the gas
Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L
Answer:
a. alkyne
b. alkane
c. alkyne
d. alkene
Explanation:
The general formula for each class of compound is given below
Alkane: 
Alkene: 
Alkyne: 
(assuming single multiple bonds)
Now let us classify according to the above formulas:
a. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
b. It has two hydrogen atoms more than the two times of carbon atoms hence, it's alkane
c. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
d. It has hydrogen atoms two times of carbon atoms hence, it's alkene
Answer: Hello i am confused are you asking a question?
Explanation:
