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tankabanditka [31]
2 years ago
11

Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Fl

ask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:
Chemistry
1 answer:
MrMuchimi2 years ago
5 0

Here is the complete question.

Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:  the following properties. (Use the notation >, <, or =, for example B=C>A.)

(a) pressure

(b) average molecular kinetic energy

(c) diffusion rate after the valve is opened

(d) total kinetic energy of the molecules

Answer:

Explanation:

Given that:

Three flask A,B, C:

contains a volume of 8-L

mass m = 4g    &;

Temperature = 276 K

Flask A = He

Flask B = H₂

Flask C = CH₄

a) From the ideal gas equation:

PV = nRT

where;

n = number of moles = mass (m)/molar mass (mm)

Then:

PV = m/mm RT

If  T ,m and V are constant for the three flasks ; then

P ∝ 1/mm

As such ; the smaller the molar mass the larger the pressure.

Now; since the molecular weight of CH₄ is greater than He and H₂ and also between He and H₂,  He has an higher molecular weight .

Then the order of pressure in the flask is :

\mathbf{P_B >P_A>P_C}

where :

P_A = pressure in the flask A

P_B = pressure in the flask B

P_C= Pressure in the flask C

b)

average molecular kinetic energy

We all know that  the average molecular kinetic energy  varies directly proportional to the temperature.

Thus; the given temperature = 276 K

∴ The order of the average molecular kinetic energy is \mathbf{K.E_A =K.E_B =K.E_C}

c)

The rate of diffusion of gas is inversely proportional to the square root of it density . Here the density is given in relation to their molar mass.

So;

rate of diffusion ∝ \dfrac{1}{\sqrt{mm} }

where;

D_A = rate of diffusion in flask A

D_B = rate of diffusion in flask B

D_C = rate of diffusion in flask C

Thus; the order of the rate of diffusion = D_B  > D_A > D_C

d)  total kinetic energy of the molecules .

The kinetic energy deals with how the speed of particles of a  substance determines how fast the substances will diffuse in a given set of condition.

The the order of the total kinetic energy depends on the molecular speed

Thus; the order of the total kinetic energy  for the three flask is as follows:

\mathbf{ K.E_B>K.E_A>K.E_C}

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concentration of unknown Cd^{2+}.

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Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. K pKa1 K pKa2 1.30
FrozenT [24]

Answer:

* Before addition of any KOH:

pH = 0,0301

*After addition of 25.0 mL KOH:

pH = 1,30

*After addition of 50.0 mL KOH:

pH = 2,87

*After addition of 75.0 mL KOH:

pH = 6,70

*After addition of 100.0 mL KOH:

pH = 10,7

Explanation:

H₃PO₃ has the following equilibriums:

H₃PO₃ ⇄ H₂PO₃⁻ H⁺

k = [H₂PO₃⁻] [H⁺] / [H₃PO₃] k = 10^-(1,30) <em>(1)</em>

H₂PO₃⁻ ⇄ HPO₃²⁻ + H⁺

k = [HPO₃²⁻] [H⁺] / [H₂PO₃⁻] k = 10^-(6,70) <em>(2)</em>

Moles of H₃PO₃ are:

0,0500L×(1,8mol/L) = 0,09 moles of H₃PO₃

* Before addition of any KOH:

Using (1), moles in equilibrium are:

H₃PO₃: 0,09-x

H₂PO₃⁻: x

H⁺: x

Replacing:

10^{-1.30} = \frac{x^2}{0.09-x}

4.51x10⁻³ - 0.050x -x² = 0

The right solution of x is:

x = 0.0466589

As volume is 0,050L

[H⁺] = 0.0466589moles / 0,050L = 0,933M

As pH = -log [H⁺]

<em>pH = 0,0301</em>

*After addition of 25.0 mL KOH:

0,025L×1,8M = 0,045 moles of KOH that reacts with H₃PO₃ thus:

KOH + H₃PO₃ → H₂PO₃⁻ + H₂O

That means moles of KOH will be the same of H₂PO₃⁻ and moles of H₃PO₃ are 0,09moles - 0,045moles = 0,045moles

Henderson-Hasselbalch formula is:

pH = pka + log₁₀ [A⁻] /[HA]

Where A⁻ is H₂PO₃⁻ and HA is H₃PO₃.

Replacing:

pH = 1,30 + log₁₀ [0,045mol] / [0,045mol]

<em>pH = 1,30</em>

*After addition of 50.0 mL KOH:

The addition of 50.0 mL KOH consume all H₃PO₃. Thus, in the solution you will have just H₂PO₃⁻. Thus, moles in solution for the equilibrium will be:

H₂PO₃⁻: 0,09-x

HPO₃²⁻: x

H⁺: x

Replacing:

10^{-6.70} = \frac{x^2}{0.09-x}

1.8x10⁻⁸ - 2x10⁻⁷x - x² = 0

The right solution of x is:

x = 0.000134064

As volume is 50,0mL + 50,0mL = 100,0mL

[H⁺] = 0.000134064moles / 0,100L = 1.34x10⁻³M

As pH = -log [H⁺]

<em>pH = 2,87</em>

*After addition of 75.0 mL KOH:

Applying Henderson-Hasselbalch formula you will have 0,045 moles of both H₂PO₃⁻ HPO₃²⁻ and pka: 6,70:

pH = 6,70 + log₁₀ [0,045mol] / [0,045mol]

<em>pH = 6,70</em>

*After addition of 100.0 mL KOH:

You will have just 0,09moles of HPO₃²⁻, the equilibrium will be:

HPO₃²⁻ + H₂O ⇄ H₂PO₃⁻ + OH⁻ with kb = kw/ka = 1x10⁻¹⁴/10^-(6,70) = 5,01x10⁻⁸

kb = [H₂PO₃⁻] [OH⁻] / [HPO₃²⁻]

Moles are:

H₂PO₃⁻: x

OH⁻: x

HPO₃²⁻: 0,09-x

Replacing:

5.01x10^{-8} = \frac{x^2}{0.09-x}

4.5x10⁻⁹ - 5.01x10⁻⁸x - x² = 0

The right solution of x is:

x = 0.000067057

As volume is 50,0mL + 100,0mL = 150,0mL

[OH⁻] = 0.000067057moles / 0,150L = 4.47x10⁻⁴M

As pH = 14-pOH; pOH = -log [OH⁻]

<em>pH = 10,7</em>

<em></em>

I hope it helps!

6 0
2 years ago
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