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ehidna [41]
3 years ago
12

Two systems with heat capacities 19.9 J mol-1 K-1 and 28.2 ] mol 1 K-1 respectively interact thermally and come to an equilibriu

m temperature of 300C. If the initial temperature of system 1 was 450C, what was the initial temperature of system 2 in °C? You may assume that the total energy of the combined systems remains constant
Chemistry
1 answer:
MAVERICK [17]3 years ago
8 0

Answer : The initial temperature of system 2 is, 19.415^oC

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

-q_1=q_2

m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)

The mass remains same.

where,

C_1 = heat capacity of system 1 = 19.9 J/mole.K

C_2 = heat capacity of system 2 = 28.2 J/mole.K

T_f = final temperature of system = 30^oC=273+30=303K

T_1 = initial temperature of system 1 = 45^oC=273+45=318K

T_2 = initial temperature of system 2 = ?

Now put all the given values in the above formula, we get

-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K

T_2=292.415K

T_2=292.415-273=19.415^oC

Therefore, the initial temperature of system 2 is, 19.415^oC

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A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f
Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

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mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

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60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.

Explanation:

Data given:

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V2  =?

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At STP the pressure is 1 atm and the temperature is 273.15 K

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V2 = \frac{91 X 75 X 273.15}{303.15 X 101.3}

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