The answer is 3.25 x 10 -3 kg is not equal to 325 cg
Explanation:
a) Both A and R are true and R is the correct
explanation of A.
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
Capillary action is defined as the ability of a liquid to go up a narrow space without the help or opposition of external forces. One of the most important factors affecting capillary action is the intermolecular forces within a substance. The higher the IMF, the greater the capillary action. The H-bonding in water gives it greater IMF than acetone, so water has greater capillary action.
