Aluminium + copper sulfate : Why is this reaction called a displacement reaction

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.

6 0

4 years ago

A low energy electromagnetic wave would also have?

swat32

The answer is low frequency and long wavelength

8 0

3 years ago

If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?

krok68 [10]

Answer:

$\large\boxed{\large\boxed{0.64M}}$

Explanation:

When you form a <em>diluted solution</em> from a mother (concentrated) solution, the moles of solute are determined by the mother solution.

The main equation is:

$Molarity=\dfrac{\text{moles of solute}}{\text{volume of the solution in liters}}$

Then, since the moles of solute is the same for both the mother solution and the diluted solution:

$\text{Molarity mother solution }\times\text{ volume mother solution}=\\\\=\text{Molarity diluted solution }\times\text{ volume diluted solution}$

Substitute and solve for the molarity of the diluted solution:

$250mL\times 0.75M=(45mL+250mL)\times M\\\\\\M=\dfrac{250mL\times 0.75M}{295mL}=0.64M$

8 0

3 years ago

onsider the reaction, NO2(g) + CO(g) → NO(g) + CO2(g), for which the rate law has been determined to be: Rate = k[NO2]2[CO]. Whi

nlexa [21]

Answer:

The correct answer is option D.

Explanation:

Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

$NO_2(g) + CO(g)\rightarrow NO(g) + CO_2(g)$

Rate of the reaction:

$R=-\frac{1}{1}\times \frac{d[NO_2]}{dt}=-\frac{1}{1}\times \frac{d[CO]}{dt}$

Rate of decrease in nitrogen dioxide concentration is equal to the rate of decrease in carbon monoxide.

Given rate expression of the reaction:

$R = k[NO2]^2[CO]$

Rate of the reaction on doubling concentration of nitrogen dioxide and carbon monoxide : R'

$R'=k(2\times [NO_2])^2(2\times [CO])=8\times k[NO2]^2[CO]=8R$

Doubling the concentrations of nitrogen dioxide and carbon monoxide simultaneously will increase the rate of the reaction by a factor of eight.

Hence, none of the given statements are true.

6 0

3 years ago

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Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.

Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0

3 years ago

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Aluminium + copper sulfate : Why is this reaction called a displacement reaction​

Aluminium + copper sulfate : Why is this reaction called a displacement reaction