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3 years ago

What stress would shift the equilibrium position of the following system to the right?

Chemistry

2 answers:

MakcuM [25]3 years ago

6 0

Answer:

Decreasing the concentration of $N_2O_3$ => To the left

Heating the system => To the left

Adding a catalyst => No change

Increasing the concentration of NO => To the left

Explanation:

We have to analyze each case:

Decreasing the concentration of N2O3 

The compound $N_2O_3$ is a reactive. When we remove the concentration of this compound the reaction has to go to the left in order to restore the initial equilibrium.

Heating the system 

In this case, the heat (energy) can be considered as a product because it has a negative sign. Therefore, if we add energy the reaction will move to the opposite side. The reactive side

Adding a catalyst

The addition of a catalyst doesn't change the equilibrium of any reaction. The catalyst addition affects the velocity of the reaction but not the equilibrium.

Increasing the concentration of NO

The NO is a product therefore if we increase the concentration of NO will displace the reaction to the opposite side. To the reactants side

yanalaym [24]3 years ago

4 0

Answer:

Decreasing the concentration of N2O3

Explanation:

This is because the products on the right of the reaction occupy more space. One (1) mole of NO and another mole of NO2 will occupy more space than the one (1) mole of N2O3. Therefore decreasing the concentration of N2O3 will shift the reaction to the right because the products will have more space to occupy – hence favoring equilibrium.

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A standard galvanic cell is constructed so that the overall cell reaction is as given below where M is an unknown metal. 2 Al3+(

jasenka [17]

Answer:

magnesium metal

Explanation:

According to the redox reaction equation, six electrons were transferred hence n=6 and F= Faraday's constant 96500C. ∆G° is given hence E°cell can easily be calculated as follows:

From ∆G°= -nFE°cell

E°cell= -∆G°/nF= -(-411×10^3/96500×6)

E°cell= 0.7098V

But for Al3+(aq)/Al(s) half cell, E°= -1.66V from standard table of reduction potentials.

E°cell= E°cathode- E°anode but Al3+(aq)/Al(s) half cell is the cathode

Hence

E°anode=E°cathode - E°cell

E°anode= -1.66-0.7098= -2.37V

This is the reduction potential of Mg hence the anode material was magnesium metal

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3 years ago

When a cold front passes, the air temperature should begin to rise true or false?

sleet_krkn [62]

False the opposite should happen!

7 0

3 years ago

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A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse

ludmilkaskok [199]

Answer:

$\large \boxed{\text{5.9 s}}$

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r \propto \dfrac{1}{\sqrt{M}}$

If you have two gases, the ratio of their rates of effusion is

$\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}$

The time for diffusion is inversely proportional to the rate.

$\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}$

Let CO₂ be Gas 1 and O₂ be Gas 2

Data:

M₁ = 44.01

M₂ = 32.00

Calculation

$\begin{array}{rcl}\dfrac{t_{2}}{t_{1}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}}\\\\\dfrac{t_{2}}{\text{5 s}}& = & \sqrt{\dfrac{44.01}{32.00}}\\\\& = & \sqrt{1.375}\\t_{2}& = & \text{5 s}\times 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}$

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3 years ago

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Olin [163]

Answer:

A. vacuum

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Zolol [24]

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