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Anna35 [415]
3 years ago
13

The first-order rate constant for the decomposition of n2o5, 2n2o5(g)→4no2(g)+o2(g) at 70∘c is 6.82×10−3 s−1. suppose we start w

ith 2.00×10−2 mol of n2o5(g) in a volume of 2.3 l . you may want to reference (page) section 14.4 while completing this problem. part a how many moles of n2o5 will remain after 4.0 min ?,
Chemistry
1 answer:
torisob [31]3 years ago
7 0

The rate constant for 1st order reaction is

K = (2.303 /t) log (A0 /A)

Where, k is rate constant

t is time in sec

A0 is initial concentration

(6.82 * 10-3) * 240 = log (0.02 /A)

1.63 = log (0.02 /A)

-1.69 – log A = 1.63

Log A = - 0.069

A = 0.82

Hence, 0.82 mol of A remain after 4 minutes.

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Is beeswax a mineral, a rock or neither
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Pretty sure its a mineral

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3. 2.15 x 3.1 x 100 =
jeka94

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666.5

Explanation:

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Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

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