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3 years ago

How many particles can be stored in a 4.0 L container at room temperature (23 C) and standard atmospheric pressure

Chemistry

1 answer:

velikii [3]3 years ago

5 0

Answer: 9.91×10²³ particles

Explanation:

To find the amount of particles, you will need to use the Ideal Gas Law with what we are given.

Ideal Gas Law: PV=nRT

After we find moles, we can use Avogadro's number to convert to particles.

$n=\frac{PV}{RT}$

P=101.3kPa=1.00 atm

V=4.0 L

T=23°C+273.15=296.15 K

R=0.08206 Latm/Kmol

$n=\frac{(1.00 atm)(4.0 L)}{(0.08206Latm/Kmol)(296.15K)}$

$n=0.164595 mol$

Now that we have moles, we can convert to particles.

$0.164595mol*\frac{6.022*10^2^3 particles}{mol} =9.91*10^2^2 particles$

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1 year ago

The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (

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Answer:

3.1 kg

Explanation:

Step 1: Write the balanced combustion equation

C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O

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The molar mass of C₈H₁₈ is 114.23 g/mol.

1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol

Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈

The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.

Step 4: Calculate the mass corresponding to 70 moles of CO₂

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2 years ago

An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +

Maurinko [17]

Answer:

The simplified expression for the fraction is $\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$

Explanation:

From the given information:

O3* → O3 (1) fluorescence

O + O2 (2) decomposition

O3* + M → O3 + M (3) deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

$\text {X} = \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) } } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{ {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) } }$

$\text {X} = \dfrac{ {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3 \times cM) }$

$\text {X} = \dfrac{ {k_3 \times cM} }{k_1 +k_2 + k_3 }$ since cM is the concentration of the inert molecule

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3 years ago