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dimaraw [331]
3 years ago
6

Two gases X and Y are found in the atmosphere in only trace amounts because they decompose quickly. When exposed to ultraviolet

light the half-life of X is 1.25 h, while that of Y is 150 min.
Suppose an atmospheric scientist studying these decompositions fills a transparent 20.0 L flask with X and Y and exposes the flask to UV light. Initially, the partial pressure of X is 25.0% greater than the partial pressure of Y.
1. Will the partial pressure of X ever fall below the partial pressure of Y? (yes or no)
2. If yes, calculate the time it take the partial pressure of X to fall below the partial pressure of Y.
Chemistry
1 answer:
igomit [66]3 years ago
5 0

Answer:

1. Yes

2. After 68.1 mins, pX < pY.

Explanation:

Assuming the total gas pressure is 1 atm, let the partial pressure of Y be y, partial pressure of X will be 0.25y + y = 1.25y

1.25y + y = 1atm

2.25y = 1 atm

y = 1 atm / 2.25 = 0.44 atm

Then partial pressure of X = 0.56 atm

The partial pressure of a gas in a mixture of gases is directly proportional to the mole fraction of the gas.

Therefore the mole fraction of X and Y is 0.56 and 0.44 respectively.

The partial pressures of X and Y becomes half of their original values at 1.25 h = 85 min and Y at 150 min respectively.

The partial pressure after some time can be calculated from the half-life equation :

m = m⁰ *  1/2ⁿ

Where m = the remaining mass, m⁰ = initial mass, and n is number of half-lives undergone.

Let partial pressures represent the mass, and n for X and Y be a and b respectively:

pX  = 0.56/2ᵃ

pY = 0.44/2ᵇ

We then determine when Partial pressure of X, pX = Partial pressure of Y, pY

0.56/2ᵃ = 0.44/2ᵇ

2ᵃ/2ᵇ = 0.56/0.44

2ᵃ/2ᵇ = 1.27

2ᵃ⁻ᵇ = 2⁰°³⁴⁵

a - b = 0.345

Let this time be t, therefore,

For X; t = 85a and For Y: t = 150b

85a = 150b

then, a = 1.76b

1.76b - b = 0.345

0.760b = 0.345

b = 0.454 and,

a = 0.345 + 0.454 = 0.799

So,  X goes through 0.779 half-lives while Y goes through 0.454 half-lives Then, the time for both X and Y to have the same amount is:

t = 150 * 0.454 = 68.1 min

After 68.1 mins, pX < pY.

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a glass container was initially charged with 1.50 mol of a gas sample at 3.75 atm and 21.7C. some of the gas was release as the
butalik [34]

Answer:

0.39 mol

Explanation:

Considering the ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

At same volume, for two situations, the above equation can be written as:-

\frac {{n_1}\times {T_1}}{P_1}=\frac {{n_2}\times {T_2}}{P_2}

Given ,  

n₁ = 1.50 mol

n₂ = ?

P₁ = 3.75 atm

P₂ = 0.998 atm

T₁ = 21.7  ºC

T₂ = 28.1 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (21.7 + 273.15) K = 294.85 K  

T₂ = (28.1 + 273.15) K = 301.25 K  

Using above equation as:

\frac{{n_1}\times {T_1}}{P_1}=\frac{{n_2}\times {T_2}}{P_2}

\frac{{1.50\ mol}\times {294.85\ K }}{3.75\ atm}=\frac{{n_2}\times {301.25\ K  }}{0.998\ atm}

n_2=\frac{{1.50}\times {294.85}\times 0.998}{3.75\times 301.25}\ mol

Solving for n₂ , we get:

n₂ = 0.39 mol

7 0
4 years ago
PLEASE HELP ME! What are some patterns or trends that are present in the table of elements?
Sati [7]

Some patterns and trend that are present in the periodic table would be

1. electronegativity (from left-to-right it increases across the table)

2. ionization (from left-to right it increases and from bottom-to-top it increases)

3. electron affinity (same as ionization energy)

4. atom radius (increases opposite way; from right-to-left it increases and from top-to-bottom it increases)

5. melting point (higher melting points with metals and lower melting point with non-metals)

6. metallic character (same as atom radius)

5 0
3 years ago
2 Zn + O2 → 2 ZnO
meriva

Answer:

we need 6.0 moles of zinc (Zn)

Explanation:

Step 1: Data given

Number of moles ZnO produced = 6.0 moles

Step 2: The balanced equation

2 Zn + O2 → 2 ZnO

For 2 moles Zinc we need 1 mol Oxygen to produce 2 moles Zinc oxide

Step 3: Calculate moles zinc

For 2 moles Zn we need 1 mol O2 to produce 2 moles ZnO

For 6.0 moles 2nO produced, we need 6.0 moles of zinc (Zn) and 3.0 moles of O2 to react.

3 0
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All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.<span>
Freezing point depression or Boiling point elevation:

</span><span>ΔT = -K (m) (i)

</span>ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent. 
<span>
K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent. 
</span><span>
m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent. 
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i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.</span>
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Umnica [9.8K]
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