c. Isoleucine has a carbon “branched” closer to the alpha carbon than does leucine.
The structure of leucine is CH3CH(<u>CH3</u>)CH2CH(NH2)COOH.
The structure of isoleucine is CH3CH2CH(<u>CH3</u>)CH(NH2)COOH.
In leucine, the CH3 group is <em>two carbons away</em> <em>from</em> the α carbon; in isoleucine, the CH3 group is on the carbon <em>next to</em> the α carbon.
Thus, <em>isoleucine</em> has the closer branched carbon.
“One is charged, the other is not” is i<em>ncorrect</em>. Both compounds are uncharged.
“One has more H-bond acceptors than the other” is <em>incorrect</em>. Each acid has two H-bond acceptors — the N in the amino and the O in the carbonyl group.
“They have different numbers of carbon atoms” is <em>incorrec</em>t. They each contain six carbon atoms.
Answer:
The answer is
<h2>32.4 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume of aluminum = 12 cm³
Density = 2.70 g/cm³
The mass of aluminum is
mass = 2.7 × 12
We have the final answer as
<h3>32.4 g</h3>
Hope this helps you
Answer:
Lattice energy is <em>the energy required to convert a mole of ionic solid into its constituent ions in the gas phase</em>
Explanation:
Lattice energy is usually calculated by the Born-Haber cycle, from the affinity energies and sublimation ethalphy values. It is used as an estimation of the ionic energy strength between the ions in an ionic compound.
It is defined as the energy needed to broke 1 mol of a given ionic compound into its ions in the gaseous state. For example, the lattice energy for sodium chloride (NaCl) is the energy required to separate 1 mol of solid ionic compound (NaCl(s)) and produce the sodium and chlorine ions in the gas phase: Na⁺(g) and Cl⁻(g).
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
