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Bezzdna [24]
3 years ago
6

What is the density of 0.50 grams of gaseous carbon stored under 1.5 atm of pressure at a temperature of -20.0 C?

Chemistry
1 answer:
Colt1911 [192]3 years ago
6 0

Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.

Explanation:

  • d = m/V, where d is the density, m is the mass and V is the volume.
  • We have the mass m = 0.50 g, so we must get the volume V.
  • To get the volume of a gas, we apply the general gas law PV = nRT

P is the pressure in atm (P = 1.5 atm)

V is the volume in L (V = ??? L)

n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).

  • Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
  • Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.
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Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

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where,

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

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

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Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is 0.449J/g^oC).

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