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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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2.50 liter of a gas has a pressure of 165. kPa at 25.0°C. If the pressure increases to 600. kPa and the temperature to 100.0°C,
tiny-mole [99]

Answer:

0.861 L

Explanation:

We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Convert the temperatures to degrees Kelvin.

25.0°C -> 298 K, 100.0°C -> 373 K

Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:

(165(2.5))/298 = (600(V₂))/373

V₂ = (165(2.5)(373))/(298(600))

V₂ = 0.861 L

7 0
3 years ago
Determine the oxidation number of sodium in Na202
Olegator [25]

Answer:

+1

Explanation:

Na₂O₂

NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.

Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:

Na₂O₂ = 0 (oxidation number of ground state compound is zero)

2Na + 2O = 0

O = –1

2Na + 2(–1) = 0

2Na – 2 = 0

Collect like terms

2Na = 0 + 2

2Na = 2

Divide both side by 2

Na = 2/2

Na = +1

Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1

5 0
3 years ago
A glow stick contains a glass vial with chemicals. when the glow stick is bent, the vial breaks and the chemicals react to produ
mina [271]

A glow stick will glow longer at lower temperatures than at room temperature, one can infer from the observation. Temperature and reaction time are the test variables.

We notice in this reaction that a glow stick stored in the freezer lights for a longer period of time than a glow stick stored at normal temperature. This implies that temperature affects how long a response lasts.

The most straightforward explanation for this observation is that glow sticks glow longer in colder temperatures than they do at room temperature; as a result, glow sticks kept in the freezer are observed to glow longer than glow sticks kept at room temperature.

To learn more about chemicals to the given link:

brainly.com/question/24600141

#SPJ4

6 0
2 years ago
What is the total energy change for the following reaction:CO+H2O-CO2+H2
Alekssandra [29.7K]

Answer:

\large \boxed{\text{-41.2 kJ/mol}}

Explanation:

Balanced equation:    CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)

(a) Enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}

(b) Total enthalpies of reactants and products

\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

4 0
3 years ago
Please help i need this like in the next 5 minutes
lana66690 [7]

it is the second one

7 0
3 years ago
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