CuCl2<span> + Fe ⟶ Cu + FeCI</span><span>2</span>
The answer is (3). The reaction that can occur at the anode is oxidation reaction which will lose electrons. So (1) and (2) are not correct. For (4) Fe3+ can not lose electrons again.
Answer:
-3135.47 kJ/mol
Explanation:
Step 1: Write the balanced equation
C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)
Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)
We will use the following expression.
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpies of formation
p: products
r: reactants
ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))
ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol
ΔH°r = -3135.47 kJ
Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.
I believe the answer is the first one <span>NaH</span>
Answer:
The aerobic cellular respiration of the glucose where glucose is converted to energy via four steps as follows
1. Glycolysis (glucose break down to pyruvic acid)
2. Link reaction
3. Krebs cycle
4. Electron transport chain, or ETC
The four pyruvic acid produces Four ATP, twenty NADH, and four
molecules
Explanation:
When four pyruvic acid enters step two of the aerobic cellular respiration, they are converted by Oxidative decarboxylation into acetyl-CoA, four molecules of NADH and four molecule of CO2 are formed. This process is otherwise called the link reaction or transition step, because it connects or links the Krebs cycle and glycolysis.
From the chemical reactions involved in cellular respiration of one glucose molecule, from two pyruvic acid molecules we have 2 ATP molecules, 10 NADH molecules, and 2 FADH2 molecules
Hence from four pyruvic acid molecules we have that the acetyl-CoA produced from the four pyruvic acid enters the the Krebs cycle and forms four ATP molecules, twenty NADH molecules, and four
molecules.