The balanced equation for the above reaction is as follows;
C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O
molar ratio of glucose to CO₂ is 1:6
the mass of glucose reacted - 24.0 g
Number of glucose moles reacted - 24.0 g /180 g/mol = 0.133 mol
Number of CO₂ formed when 0.133 mol of glucose reacted -0.133 x 6 = 0.798 mol
since CO₂ is a gas we can use the ideal gas law equation to find the volume occupied
PV = nRT
where P - pressure - 0.960 atm x 101 325 Pa / atm = 97 272 Pa
n - number of moles - 0.798 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 273 + 37 °C = 310 K
substituting the values in the equation
97 272 Pa x V= 0.798 mol x 8.314 Jmol⁻¹K⁻¹ x 310 K
V = 21.1 dm³
volume occupied by CO₂ is 21.1 dm³
Answer:
361 K
Explanation:
From the tables:
Given that:
From the charts, at 
Interpolating, we get 
, T = 361 K
[ H₃O⁺] = 10 ^ - pH
[ H₃O⁺ ] = 10 ^ - 7.30
[ H₃O⁺ ] = 5.011 x 10⁻⁸ M
hope this helps!
Vegetative propagation is a form of asexual reproduction of a plant. Only one plant is involved and the offspring is the result of one parent. The new plant is genetically identical to the parent. i hope this helped <3
