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professor190 [17]
3 years ago
8

Ethanol (c2h5oh) melts at -114°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9

7 j/gk and 2.3 j/gk, respectively. how much heat (kj) is needed to convert 25.0 g of solid ethanol at -135°c to liquid ethanol at -50°c?
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
6 0

Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

Explanation:

Temperature of Solid C_2H_5OH=-135^oC=138 K(0^oC=273K)

Melting temperature of Solid C_2H_5OH=114^oC=159 K

Temperature of liquid C_2H_5OH=-50^oC=223K

Specific heats of solid  ethanol = 0.97 J/gK

Specific heats of liquid ethanol = 2.3 J/gK

Heat required to melt the the 25 g solid C_2H_5OH at 159 K

\Delta T_1 = 159 K - 138 K = 21 K

Q_1=mc\Delta T= 25\times 0.97J/gK\times 21 K=509.25 J

Heat required to melt and raise the temperature of C_2H_5OH upto 223 K

\Delta T_2 = 223 K - 159 K  = 64 K

Q_2=mc\Delta T= 25\times 2.3J/gK\times 64 K=3680 J

Total heat to convert solid ethanol to liquid ethanol at given temperature :

Q_1+Q_2=509.25 J+3680 J=4189.25 J=4.18925 kJ (1kJ=1000J)

Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

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A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal
devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0
3 years ago
Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).
mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

Mass of CuSO₄.5H₂O = ?

To solve this problem, we must write the chemical relationship between both species.;

             CuSO₄.5H₂O  →   CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

         Number of moles  = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

           Volume  = 250 x 10⁻³L

  Input the parameters and solve for number of moles;

        Number of moles  = 250 x 10⁻³  x  2.01 = 0.5mol

From the equation;

             1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O  

So  0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16)  = 249.6g/mole

Mass of CuSO₄.5H₂O = number of moles x molar mass

                                      = 0.5 x 249.6

                                     = 124.8g

The mass of CuSO₄.5H₂O is 124.8g

5 0
3 years ago
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