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Ket [755]
2 years ago
8

If there are 3 moles of gas A, 4 moles of gas b and 5 moles of gas c I'm a mixture of gases and the pressure A is found to be 2.

5 atm, what is the total pressure of the sample of gases
Chemistry
1 answer:
Vlad1618 [11]2 years ago
4 0

Answer:

Total pressure = 10 atm

Explanation:

The excersise can help to understant the concept of mole fraction

Mole fraction is defined as:

moles of gas / Total moles = pressure of that gas / Total pressure

Both can be summed.

Sum of moles of each gas in the mixture = Total moles

Sum of pressures of each gas in the mixture = Total pressure

Those, are laws for gases.

Total moles: 3 mol of gas A + 4 moles of gas B + 5 moles of C

                 = 12 total moles

Pressure of A is 2.5 atm.

We replace data at the mole fraction relation

moles of A / total moles = Pressure of A  / Total pressure

3 moles / 12 moles = 2.5 atm / Total pressure

Total pressure = 2.5 atm / (3mol/12mol)

Total pressure = 10 atm

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1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation
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Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

<em>Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:</em>

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

<em>Q2:  Calculate the energy change (q) of the surroundings (water) using the enthalpy equation </em>

<em>qwater = m × c × ΔT.  </em>

<em>We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.</em>

<em></em>

a) First part: the energy change (q) of the surroundings (water):

  • The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (25 g)(4.18 J/g°C)(6.4°C) = <em>668.8 J.</em>

<em>b) second part:</em>

<em>Q water = Q unknown metal. </em>

<em>Q unknown metal =  - </em>668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

<em>Q unknown metal =  - </em>668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

<em>- </em>668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (<em>- </em>668.8)/(- 1914) = 0.3495 J/g°C.

<em></em>

3 0
3 years ago
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