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4 years ago

Which description refers to fog?

Chemistry

1 answer:

Aneli [31]4 years ago

4 0

Answer:

The correct option is;

Is found near bodies of water

Explanation:

Fog forms as an aerosol which is visible to the eye that normally consists of tiny condensed water vapor droplets or crystals of ice that is formed when air that is warm comes in contact with cool air close to the Earth's surface. Fog can be taken as a stratus like cloud that lies close to the Earth's surface and it normally forms due to the presence of water bodies that are nearby, the topography of the area or the conditions of the wind.

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3 years ago

Respuesta funcion oxido de Mg+2 + O-2

Effectus [21]

Answer:

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Explanation:

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3 years ago

The dissociation of a/an ________ releases hydrogen ions and increases the concentration of hydrogen ions in a solution.

Crank

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3 years ago

3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con

KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

$m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}$

c) Half life for first-order kinetics is computed by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min$

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

$C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M$

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3 years ago