Answer: ORGANIC ACIDS
Explanation:
CAM PLANTS CARBOXYLATE ORGANICS ACIDS through the addition of CO2 to PEP Carboxylase( a phosphoenolpyruvate carboxylase enzyme present in the mesophyll cells of the cytoplasm in a green plant) to produce Oxaloacetate (organic compound).
CO2 + PEP ⇒ C4H4O5 (oxaloacetate)
Oxaloacetate is then converted to a similar molecule, Malate (C4H6O5, another form of organic compound) that can be transported in to the bundle-sheath cells. Malate enters the plasmodesmata and releases the CO2. The CO2 then fixed by rubisco and made into sugars via the Calvin cycle.
Answer:
T2 = 94.6 C
Explanation:
Use Clausius-Clayperyon equation.
ln P1/P2 = ∆Hvap/R (1/T2 - 1/T1) where R = 8.314 J/mol-K and T is in degrees K
P1 = 760 mmHg
P2 = 630 mmHg
T1 = 373 K
T2 = ?
∆Hvap = 40.7 kJ/mole
R = 0.008314 kJ/mole-K (NOTE: change R to units of kJ)
Plug in and solve for T2
ln 760 mmHg/630 mmHg = 40.7 kJ/mole (1/T2 - 1/373K)
T2 = 367.74 K = 94.6 C
Answer:
Kc = [C2H5OH]/{[C2H4][H2O]}
[H2O] = [C2H5OH]/{[C2H4] x Kc}
[H2O] = 1.69/(0.015 x 9.0 x 10^3) = 0.013 M
Explanation:
Kc is the equilibrium constant, it is egal to the product of the equilibrium concentration of the product dived by the product of the equilibrium concentration of the reactants.
with that formula and what is given, male the equilibrium concentration of H2O the subject of the formula and calculate its concentration by substituting given values.
A conjugate base is the species that D. Remains after an acid has given up a single proton.
It is the deprotonated form of the acid.
