Answer:
QED, EDQ, DQE, EQD, DEQ, QDE
Explanation:
The structure of all possible peptides that contain the given amino acids are :
QED, EDQ, DQE, EQD, DEQ, QDE
where : Asp is represented by the letter code D
Glu is represented by the letter code E
Gln is represented by the letter code Q
Note : when three amino acids combine they form what is known as tripeptide ( i.e. contains two peptide linkages ) while a peptide linkage is been formed by the combination of a carboxyl group of an amino acid and the amino group of different amino acid
Answer:
0.0187 M
Explanation:
Step 1: Write the balanced neutralization reaction
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
18.7 mL of 0.01500 M HCl react.
0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol
Step 3: Calculate the reacting moles of NaOH
The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.
Step 4: Calculate the molarity of NaOH
2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.
[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M
Answer:
D. The coefficients give the ratio of mole reactant to moles product.
Explanation:
In stoichiometric calculations, the amount of product formed from reactants can be determined.
- Using this approach, the number of moles of reactants and products on both sides of the expression must be balanced.
- As a rule of thumb, the coefficients give the ratio of moles of reactants to moles of products.
- This is very useful in a number of calculations using the stoichiometric approach.
Answer:
-3
Explanation:
The oxidation state or oxidation number of an atom is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.
The complex anion here is [Cr(CN)6]3-.
Now, as the oxidation state of CN or cyanide ligand is -1, and if we suppose the oxidation state of Cr to be 'x', then; x - 6 = -3 (overall charge on the anion),
so x= +3. Hence the oxidation state of Chromium in this complex hexacyanochromium (III) anion comes out to be -3.
.
Answer:
84.75°C is the boiling point of water at an elevation of 7000 meter.
Explanation:
Rate of change of pressure = 19.8 mmHg/1000 ft
1 foot = 
meter
Pressure change for every 1 m = 0.065 mmHg × 1= 0.065 mmHg
Elevation Pressure
0 m 760 mmHg
1000 m 695 mmHg
2000 m 630 mmHg
Pressure drop at the elevation of 7000 m: 
Pressure at 7000 m = 760 mmHg - 455mmHg = 305 mmHg
The boiling point of water decreases 0.05°C for every 1 mmHg drop in atmospheric pressure.
At 7000 meter elevation the boiling of water will be :
Boiling point of water at 7000 meter elevation :
84.75°C is the boiling point of water at an elevation of 7000 meter.
