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a_sh-v [17]
3 years ago
10

The rate constant, k, for a first-order reaction is equal to 4.3 × 10-4 s-1. What is the half-life for the reaction? The rate co

nstant, k, for a first-order reaction is equal to 4.3 × 10-4 s-1. What is the half-life for the reaction? 1.9 × 103 s 1.2 × 103 s 1.6 × 103 s 3.0 × 10-4 s
Chemistry
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

1.6 × 10³ s

Explanation:

Let's consider the following generic reaction.

A → B

The rate law is:

rate=k \times [A]^{m}

where,

rate is the reaction rate

k is the rate constant

[A] is the molar concentration of the reactant A

m is the reaction order

When m = 1, we have a first-order reaction. We can calculate the half-life for this reaction using the following expression.

t_{1/2}=\frac{ln2}{k} =\frac{ln2}{4.3 \times 10^{-4}s^{-1} }=1.6 \times 10^{3}s

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Assume the volume of the crew cabin is 74,000 L. The pressure is maintained at around 1.00 atm, ideally with an 80% nitrogen and
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Answer:

The number of moles of oxygen present in the crew cabin at any given time is 615.309 moles

Explanation:

The given parameters are;

Volume of the crew cabin = 74,000 L

Pressure of the crew cabin = 1.00 atm

Percentage of nitrogen in the mixture of gases in the cabin = 80%

Percentage of oxygen in the mixture of gases in the cabin = 20%

Temperature of the cabin = 20°C = 293.15 K

Therefore, volume of oxygen in the crew cabin = 20% of 74,000 L

Hence, volume of oxygen in the crew cabin = \frac{20}{100} \times 74,000 \, L = 14,800 \, L

From the universal gas equation, we have;

n = \frac{P \times V}{R  \times  T}

Where:

n = Number of moles  of oxygen

P = Pressure = 1.00 atm

V = Volume of oxygen = 14,800 L

T = Temperature = 293.15 K

R = Universal Gas Constant = 0.08205 L·atm/(mol·K)

Plugging in the values, we have;

n = \frac{1 \times 14,800 }{0.08205   \times  293.15 } = 615.309 \, moles

The number of moles of oxygen present in the crew cabin at any given time = 615.309 moles.

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Which functional group is found in methanol?
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