Since glycolysis of one glucose molecule generates two acetyl CoA molecules, the reactions in the glycolytic pathway and citric acid cycle produce six CO2 molecules, 10 NADH molecules, and two FADH2 molecules per glucose molecule
Answer:
232.5 g C2H6O2
Explanation:
The equation you need to use here is ΔTf = i Kf m
Since pure water freezes at 0 C, your ΔTf is just 4.46 C
i = 1 (ethylene glycol is a weak electrolyte)
Kf = molal freezing constant, which for water is 1.86 C/m
m = molality = x mols C2H6O2 / 1.15 kg H2O (don't know the moles of ethylene glycol we're dissolving yet)
Than,
4.46 C = 1.86 C/m (x mol C2H6O2 / 1.15 kg H2O)
Solve for x, you should get x = 2.75 mol C2H6O2
3.75 mol C2H6O2 (62 g C2H6O2 / 1 mol C2H6O2) = 232.5 g C2H6O2
Answer:
0.2 M.
Explanation:
- For the acid-base neutralization, we have the role:
The no. of millimoles of acid is equal to that of the base at the neutralization.
<em>∴ (XMV) KOH = (XMV) H₂SO₄.</em>
X is the no. of reproducible H⁺ (for acid) or OH⁻ (for base),
M is the molarity.
V is the volume.
X = 1, M = 0.5 M, V = 38.74 mL.
X = 2, M = ??? M, V = 50.0 mL.
∴ M of H₂SO₄ = (XMV) KOH/(XV) H₂SO₄ = (1)(0.5 M)(38.74 mL)/(2)(50.0 mL) = 0.1937 M ≅ 0.2 M.
Answer:
0.049 mol/L.s
Explanation:
The decomposition of hydrogen peroxide is:
![Rate = -\dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{ 2 \Delta [H_2O_2]}{\Delta t}](https://tex.z-dn.net/?f=Rate%20%3D%20-%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%202%20%20%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D)
The rate of decomposition reaction = the rate of formation of 
= 0.098 mol/L.s
∴
Rate of formation of 
= 0.049 mol/L.s
