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Kazeer [188]
3 years ago
6

Changes in pressure can have a large effect on equilibrium systems containing gaseous components.

Chemistry
2 answers:
Paha777 [63]3 years ago
5 0

Answer:

1. Increase volume.

2. No change.

3. No change.

Explanation:

Hello,

In this case, if we want to shift the reaction rightwards, based on the Le Chatelier's principle we would have to:

1. For this reaction:

CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)

- Increase the volume or decrease the pressure, since there are more gaseous moles at the products.

2. For this reaction:

S(s) + 3F_2(g) \rightleftharpoons SF_6(g)

- Do nothing since it is not possible to achieve it as we have the same number of gaseous moles at both reactants and products.

3. For this reaction:

Cl_2(g) + I_2(g)\rightleftharpoons 2ICl(g)

- Do nothing since it is not possible to achieve it as we have the same number of gaseous moles at both reactants and products.

Regards.

gayaneshka [121]3 years ago
3 0

Answer:

The correct option is 1, since by changing the partial pressures the gas pressures change, the gases go from the zones of higher partial pressure to the zones of lower partial pressure, an example of this is the homeostasis of the human pulmonary alveolus in gas exchange with CO2 and O2.

Explanation:

In the first it increases, in the second the volume is maintained, and in the third reaction it decreases.

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Liquids have free and fast moving particles. Liquids do not take a specific shape unless they are enclosed such as in a bottle. Anyway, since they are free flowing when they freeze, their movement becomes very slow. Eventually the particles will not have room to move as they used to and will be held in one place. The particles are still moving but they are only shaking or vibrating in the same place. You can't even tell this movement is happening, and that is what happens when liquid freezes.

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If you heat your melting point tube too quickly, what will your observed melting point be in comparison to the actual melting po
solniwko [45]

Explanation:

When conducting a melting point experiment, if we were to heat a sample quickly. Large amount heat is provided instantly which would melt the crystals in the tube very quickly, even before the temperature of the thermometer reaches to that level. So the observes melting point would be much lower than the actual melting point when sample is heated slowly.

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The rate constant for this first‑order reaction is 0.550 s−10.550 s−1 at 400 ∘C.400 ∘C. A⟶products A⟶products How long, in secon
Alisiya [41]

Answer:

t=2.08s

Explanation:

Hello,

In this case, for first order reactions, we can use the following integrated rate law:

ln(\frac{[A]}{[A]_0} )=kt

Thus, we compute the time as shown below:

t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s

Best regards.

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