2 years ago

Which of these compounds would you expect

Chemistry

1 answer:

Veseljchak [2.6K]2 years ago

6 0

ClBr, two nonmetals

Hope this helps you

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What is true of a synthesis reaction?

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There is only one product in a synthesis reaction.

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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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3 years ago

How many molecules units are in 4.5 moles of CF4

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1.0 mole ---------- 6.02x10²³ molecules
4.5 moles -------- ?

4.5 * 6,02x10²³ / 1.0

= 2.709x10²⁴ molecules units

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3 years ago

A sample of an unknown liquid has a volume of 24.0 mL and a mass of 6 g. What is its density? Show your work or explain how you

kobusy [5.1K]

The answer would be 0.25 g/mL.
I determined the density by dividing the mass by the volume which gives you the density. D = mass/volume.
<span>6 g / 24 mL = 0.25 g/mL </span>

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2 years ago

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