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3 years ago

A 0.316 mol sample of nitrogen gas, N2 (g), is placed in a 4.00 L container at 315 K.

1 answer:

7 0

2.04 atm is the pressure of the nitrogen gas in the container when 0.316 mol sample of nitrogen gas, N2 (g), is placed in a 4.00 L container at 315 K.

Explanation:

Data given:

moles of nitrogen gas, n = 0.316

volume of the nitrogen gas, V = 4 litres

temperature of the container of nitrogen gas = 315 K

R (gas constant) = 0.08201 Latm/mole K

Pressure of the nitrogen gas on the container, P = ?

from the data given, we will apply ideal gas law equation to calculate the pressure on the nitrogen gas:

PV = nRT

Rearranging the equation,

P = $\frac{nRT}{V}$

P = $\frac{0.316 X 0.08201 X 315}{4}$

P = 2.04 atm

The pressure of the nitrogen gas in the container is 2.04 atm.

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The rate law for 2NO(g) + O2(g) ====>2NO2(g) is rate=k[NO]^2[O2]. In addition to the mechanism, the following have been propo

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Answer:

Mechanism (I).

Explanation:

The given chemical reaction:

2NO(g) + O₂(g) → 2NO₂(g)

The given rate equation for the reaction:

rate = k [NO]² [O₂]

The rate equation of a given chemical reaction is equal to the product of the concentration of reactants raised to the power the partial orders of reaction.

The overall rate of a given multi-step chemical reaction depends on the slowest step in the reaction mechanism.

Therefore, the slowest step of a given chemical reaction is called the rate-determining step (RDS).

The proposed reaction mechanisms:

(I). 2NO(g) + O₂(g) → 2NO₂(g)

In an elementary (single step) reaction, the rate equation can be written as:

rate = k [NO]² [O₂]

(II). 2NO (g) ===> N₂O₂ (g) [fast]

N₂O₂(g) +O₂(g) ===> 2NO₂ (g) [slow] ⇒ RDS

In an complex (multi-step) reaction, the slow step is the rate determining step (RDS). So, the rate equation can be written as:

rate = k [N₂O₂] [O₂]

(III). 2NO(g) ===>N₂(g) + O₂(g) [fast]

N₂(g) + 2O₂(g) ===> 2NO₂(g) [slow] ⇒ RDS

In an complex (multi-step) reaction, the slow step is the rate determining step (RDS). So, the rate equation can be written as:

rate = k [N₂] [O₂]²

Thus, the rate equation of the reaction mechanism: (I). is the same as the given rate law of the chemical reaction.

Therefore, reaction mechanism (I). is consistent with the rate law of the given chemical reaction.

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4 years ago

How many moles of H20 are contained in 4.21 x 1024 molecules of H2O?

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Answer:

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Explanation:

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Number of moles : 4.21 x 10^24 / 6.022 x 10²³ = 6.99 moles ( rounded )

The answer is C

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3 years ago