The lowest pressure attainable using the best available vacuum techniques is about 10−12 n/m2. part a at such a pressure, how many molecules are there per cm3 at 2 ∘c?
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Ответ:
Number of molecules: N = 263.31
Further explanationSome of the laws regarding gas can apply to ideal gas (volume expansion does not occur when the gas is heated),:
Boyle's law at constant T,![\displaystyle P = \dfrac {1} {V}](/tpl/images/0079/6058/1cf90.png)
Charles's law, at constant P,![\displaystyle V = T](/tpl/images/0079/6058/60071.png)
Avogadro's law, at constant P and T,So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written
where
P = pressure, atm , N/m²
V = volume, liter
n = number of moles
R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)
T = temperature, Kelvin
n = N / No
n = mole
No = Avogadro number (6.02.10²³)
n = m / m
m = mass
M = relative molecular mass
Known
P = 10−12 N / m2
V = 1 cm3 = 10-6 m3
T = 2 ºC = 2 + 273 = 275 K
R = 8,314 J / mol. K
then the number of molecules (N):
N = n x No
N = 4,374.10⁻²² x 6.02.10²³
N = 263.31
Learn moreWhich equation agrees with the ideal gas law
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Which law relates to the ideal gas law
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Ответ:
25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S
500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.
650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃
Explanation:
When 2 compounds that produce an insoluble substance are mixed together, A precipitate will be formed if Q of reaction > Ksp
For the solutions:
1.5 L of 0.025 M BaCl₂ and 1.25L of 0.014 M Pb(NO₃)₂.
Ksp is:
PbCl₂(s) ⇄ Pb²⁺(aq) + 2Cl⁻(aq)
Ksp = 2.4x10⁻⁴ = [Pb²⁺][Cl⁻]²
Molar concentration of each ion is:
[Pb²⁺] = 1.25L ₓ (0.014mol / L) = 0.0175mol / 2.75L = 6.36x10⁻³M
[Cl⁻] = 2 ₓ 1.5L ₓ (0.025mol / L) = 0.075mol / 2.75L = 0.0273M
Replacing in Ksp expression to find Q:
Q = [6.36x10⁻³M][0.0273M]² = 4.73x10⁻⁶
As Q < Ksp, the mixture will not produce a precipitate.
25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S
Ksp is:
CoS(s) ⇄ Co²⁺(aq) + S²⁻(aq)
Ksp = 4.0x10⁻²¹ = [Co²⁺][S²⁻]
Molar concentration of each ion is:
[Co²⁺] = 0.025L ₓ (1x10⁻⁵mol / L) = 2.5x10⁻⁷mol / 0.1L = 2.5x10⁻⁶M
[S²⁻] = 0.075L ₓ (5x10⁻⁴mol / L) = 3.75x10⁻⁵mol / 0.1L = 3.75x10⁻⁴M
Replacing in Ksp expression to find Q:
Q = [2.5x10⁻⁶M][3.75x10⁻⁴M] = 9.38x10⁻⁶
As Q > Ksp, the mixture will produce a precipitate.
500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.
Ksp is:
Hg₂Cl₂(s) ⇄ 2Hg⁺(aq) + 2Cl⁻(aq)
Ksp = 3.5x10⁻¹⁸ = [Hg⁺]²[Cl⁻]²
Molar concentration of each ion is:
[Hg⁺] = 2ₓ0.100L ₓ (1.7x10⁻⁵mol / L) = 3.4x10⁻⁶mol / 0.6L = 5.67x10⁻⁶M
[Cl⁻] = 3 ₓ 0.500L ₓ (7.5x10⁻⁴mol / L) = 1.125x10⁻³mol / 0.6L = 1.88x10⁻³M
Replacing in Ksp expression to find Q:
Q = [5.67x10⁻⁶M]²[1.88x10⁻³M]² = 1.14x10⁻⁶
As Q > Ksp, the reaction will produce a precipitate.
650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃
Ksp is:
Ag₂SO₄(s) ⇄ 2Ag⁺(aq) + SO₄²⁻(aq)
Ksp = 1.5x10⁻⁵ = [Ag⁺]²[SO₄²⁻]
Molar concentration of each ion is:
[Ag⁺] = 0.175L ₓ (0.15mol / L) = 0.02625mol / 0.825L = 0.0318M
[SO₄²⁻] = 0.650L ₓ (0.080mol / L) = 0.052mol / 0.825L = 0.0630M
Replacing in Ksp expression to find Q:
Q = [0.0318M]²[0.0630M] = 6.37x10⁻⁵
As Q > Ksp, the reaction will produce a precipitate.