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Vinil7 [7]
3 years ago
9

Which of the following is true of evaporation? A. It is considered a cooling process and it raises the average kinetic energy of

the liquid because the higher energy molecules turn to vapor. B. It is considered a heating process and it lowers the average kinetic energy of the liquid because the higher energy molecules turn to vapor. C. It is considered a cooling process and it lowers the average kinetic energy of the liquid because the higher energy molecules turn to vapor. D. It is considered a heating process and it lowers the average kinetic energy of the liquid because the lower energy molecules turn to vapor.
Chemistry
1 answer:
True [87]3 years ago
4 0

Answer:

C

Explanation:

C. It is considered a cooling process and it lowers the average kinetic energy of the liquid because the higher energy molecules turn to vapor.

Moreso, evaporation does not increase the average kinetic energy because it is a surface phenomenon. Also, as molecules excape at a fast moving rate, the left over molecules have a lower kinetic energy and this consequently decreases the temperature of the liquid.

When molecules absorbs heat, the evaporate and this phenomental process provides cooling in the system

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Delta H equals 98.8 KJ, and Delta S equals 141.5 J/K. Is this reaction spontaneous or nonspontaneous at high and low pressure
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Which of the following reactions will have the largest equilibrium constant (K) at 298 K? Which of the following reactions will
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Answer:

<em>2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ </em>

Explanation:

If we know the ΔG° of a chemical reaction it is possible to calculate the equilibrium constant (k) of this procedure with the next equation:

ln Keq = -ΔG° / RT (1)

Where: Keq is equilibrium contant, ΔG° is standard state free energy change, R is gas constant and T is temperature.

Watching (1), it is possible to know that the large negative ΔG° the largest equilibrium constant. That is because R and T are always positive and to cancel the negative of equation it is necessary that ΔG° be negative.

Knowing this, is the oxidation of Hg the reaction that has the largely negative ΔG°. So, this reaction will have the largest equilibrium constant.

<em>2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ </em>

CaCO₃(s) → CaO(s) + CO₂(g) ΔG° =+131.1 kJ

3 O₂(g) → 2 O₃(g) ΔG° = +326 kJ

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I hope it helps!

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3 years ago
Which of the following most accurately represents John Dalton's model of the atom
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8 0
3 years ago
A sample of the sugar d-ribose (C5H10O5) of mass 0.727 g was placed in a calorimeter and then ignited in the presence of excess
alexandr1967 [171]

Answer:

The internal energy of combustion of d-ribose = -2127 kJ/mol

The enthalpy of formation of d-ribose = -1269.65 kJ/mol

Explanation:

Step 1: Data given

Mass of d-ribose = 0.727 grams

The temperature rose by 0.910 K

In a separate experiment in the same calorimeter, the combustion of 0.825 g of benzoic acid, for which the internal energy of combustion is −3251 kJ mol−1, gave a temperature rise of 1.940 K.

Molar mass of benzoic acid = 122.12 g/mol

Step 2: Calculate ΔU  for benzoic acid

The calorimeter is a constant-volume instrument so:

ΔU = q

ΔU = (0.825 g/ 122.12 g/mol) * (−3251 kJ /mol)

ΔU = -21.96 kJ

Step 3: Calculate ΔU  for d-ribose

c = |q| / ΔT

⇒ with ΔT = 1.940 K

c = 21.96 kJ / 1.940 K

c = 11.32 kJ /K

For d-ribose: ΔU = -cΔT

ΔU  = -11.32 kJ/K * 0.910 K

ΔU = - 10.3 kJ

Step 4: Calculate moles of d-ribose

moles ribose = 0.727 grams / 150.13 g/mol

moles ribose = 0.00484 moles

Step 5: Calculate the internal energy of combustion for d-ribose

ΔrU = ΔU / n

ΔrU  = -10.3 kJ / 0.004842 moles

ΔrU = -2127 kJ/mol

Step 6: Calculate The enthalpy of formation of d-ribose

The combustion of ribose is:

C5H10O5(s) + 5O2(g) → 5CO2(g) + 5H20(l)

Since there is no change in the number of moles of gas,  ΔrH = ΔU  

For the combustion of ribose, we consider the following reactions:

5CO2(g) + 5H2O(l) → C5H10O5(s) +5O2(g)      ΔH = -2127 kJ/mol

C(s) + O2(g) → CO2(g)      ΔH = -393.5 kJ/mol

H2(g) + 1/2 O2(g) → H2O(l)    ΔH = -285.83 kJ/mol

ΔH = 2127 kJ/mol + 5(-393.5 kJ/mol) + 5(-285.83 kJ/mol)

ΔH = 2127 kJ/mol - -1967.5 kJ/mol - 1429.15 kJ/mol

ΔH =  -1269.65 kJ/mol

The internal energy of combustion of d-ribose = -2127 kJ/mol

The enthalpy of formation of d-ribose = -1269.65 kJ/mol

3 0
3 years ago
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