Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.
Explanation:
1) Molarity of 0.250 L HCl solution : 0.0328 M
Moles of HCl in 0.250 L solution = 0.0082 moles
2) Molarity of 0.100 L NaOH solution : 0.0245 M
Moles of NaOH in 0.100 L solution = 0.00245 moles
3) Concentration of hydrochloric acid in the resulting solution.
0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.
Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L
Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles = 0.00575 moles
Molarity of HCl left un-neutralized :
0.0164 molar concentration of hydrochloric acid in the resulting solution.
Q1)
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M
Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M
Q3)
Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M
Mass = mass/molar mass of ch3ch2nh2
the molar mass of CH3CH2NH2 = 12 +(1x3)+12+(1 x2)+14+(1x2) =45 g/mol
moles is therefore = 1.50g /45g/mol = 0.033 moles
Answer:
Zirconium tetrafluoride has 4 atoms of flourine and 1 atom of Zirconium
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-)).
