Answer:
see explanation below
Explanation:
To do this exercise, we need to use the following expression:
P = nRT/V
This is the equation for an ideal gas. so, we have the temperature of 22 °C, R is the gas constant which is 0.082 L atm / mol K, V is the volume in this case, 5 L, and n is the moles, which we do not have, but we can calculate it.
For the case of the oxygen (AW = 16 g/mol):
n = 30.6 / 32 = 0.956 moles
For the case of helium (AW = 4 g/mol)_
n = 15.2 / 4 = 3.8 moles
Now that we have the moles, let's calculate the pressures:
P1 = 0.956 * 0.082 * 295 / 5
P1 = 4.63 atm
P2 = 3.8 * 0.082 * 295 / 5
P2 = 18.38 atm
Finally the total pressure:
Pt = 4.63 + 18.38
Pt = 23.01 atm
The element used in advertising signs is neon.
Answer:
(A) pH < 1 the predominant form is the cation: H3C-C(H)(NH3+)-COOH
(B) pH = pl the predominant form is the zwitterion H3C-C(H)(NH3+)-COO-
(C) pH > 11 the predominant form is the anion: H3C-C(H)(NH2)-COO-
(D) Does not occurs in any significant pH: H3C-C(H)(NH2)-COOH
Explanation:
Amino acids are bifunctional because they have an amine group and a carboxyl group. The amine group is a weak base and the carboxyl group is a weak acid, but the pKa of both groups will depend on the whole structure of the amino acid. Also, every amino acid has an isoelectric point (pI), which means the pH were the predominant form of the amino acid is the zwitterion. The structure of the alanine (CH3CH2NH2COOH) shows it has the carboxyl group at C1 with a pKa1 of 2.3 and the amino group at C2 whit the pKa2 of 9.7. The isoelectric poin (pI) of Alanine is 6. Consequently, the protonation of the molecule will depend on the pH of the solution. There are three possibilities:
1) If the pH is under the pKa of the carboxyl group (2.3) the predominant form will be with the amino group protonated, forming a cation (CH3CH(NH3+)COOH).
2) If the pH is between pKa1 (2.3) and pKa2 (9.7) the predominant form will be the zwitterion (CH3CH(NH3+)(COO-)).
3) If the pH is upper the pKa2 of the amino group (9.7) the predominant form will be with the carboxyl group deprotonated, forming an anion (CH3CHNH2(COO-)).
Answer:
-86.02 kJ/ mole
Explanation:
The moles of the acid used = Molarity × Volume (L) =
= 0.50 (0.0372 L)
= 0.0186 moles
The heat released = -1.6 kJ
∴ 0.0186 moles neutralization of HA heat is: -1.6 kJ
The molar heat of neutralization due to one mole of the unknown acid = -1.6/0.0186
= -86.02 kJ/ mole
Answer:
The mass is recorded as 32.075 g
Explanation:
"The first digit of uncertainty is taken as the last significant digit", this is the rule for significant figures in the analysis. The balance measures the mass up to three decimal places, so it makes the most sense to note the whole figure.
