A decrease in the overall volume of gases namely hydrogen would prevent nuclear fusion in a nebula.
Answer:
V = 6.17 L
Explanation:
Given data:
Volume = ?
Number of moles = 0.382 mol
Pressure = 1.50 atm
Temperature = 295 k
R = 0.0821 L. atm. /mol. k
Solution:
According to ideal gas equation:
PV= nRT
V = nRT/P
V = 0.382 mol × 0.0821 L. atm. /mol. k ×295 k / 1.50 atm
V = 9.252 L. atm. / 1.50 atm
V = 6.17 L
H₂CO₃ ⇔ HCO₃⁻ + H⁺
I 0.160 0 0
C -x +x +x
E 0.160-x +x +x
Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]
4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)
x² = 6.88 x 10⁻⁸
x = 2.62 x 10⁻⁴
HCO₃⁻ ⇔ CO₃⁻² + H⁺
I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴
C -x +x +x
E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x
Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]
5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)
x = 5.6 x 10⁻¹¹
Thus,
[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M
[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M
[CO₃⁻²] = 5.6 x 10⁻¹¹ M
[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M
[OH⁻] = 3.8 x 10⁻¹¹
Answer:
The correct statements that you must check are:
- The oxygen atom has a greater attraction for electrons than the hydrogen atom does (second statement).
- The electrons of the covalent bond are not shared equally between the hydrogen and oxygen atoms (fourth statement).
Explanation:
Electronegativity is the relative ability of an atom to pull the electrons in a covalent bond.
Hydrogen has an electronegativity of 2.20 and oxygen has 3.44. That means that oxygen attracts the electrons more strongly than hydrogen does (second statement).
As consequence, the electrons in the covalent bond H - O of water are not shared equally (fourth statement): the electron density will be higher around the O atoms.
Of course, this discards the statement telling that hydrogen atom attracts electrons much more strongly than the oxygen atom, and the statement telling that hydrogen and oxigen have same electronegativity.
Such difference in electron densities creates a dipole moment, so you discard the last statement (that the water dipole moment is equal to zero).
