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Snezhnost [94]
2 years ago
11

What is the mass, in g, of 0.500 mol of 1,2-dibromoethane, CH2BrCH2Br?

Chemistry
1 answer:
inna [77]2 years ago
4 0
Molar mass <span>CH2BrCH2Br = 188.0 g/mol

1 mole ---------- 188.0 g
</span>0.500 moles ----- ?

mass = 0.500 * 188.0 / 1

= 94.0 g

Answer C

hope this helps!
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A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water
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Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

Q_{water} -Q_{sample} = 0

Q_{water} = Q_{sample}

Where:

Q_{water} - Heat received by water, measured in joules.

Q_{sample} - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)

Where:

m_{s}, m_{w} - Mass of water and the sample of aluminium, measured in kilograms.

c_{p,s}, c_{p,w} - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

T_{s}, T_{w} - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

T - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}

T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}

Given that m_{s} = 0.380\,kg, m_{w} = 2.40\,kg, c_{p,s} = 910\,\frac{J}{kg\cdot K}, c_{p,w} = 4186\,\frac{J}{kg\cdot K}, T_{s} = 378\,K and T_{w} = 273\,K, the final temperature of the system is:

T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}

T = 276.494\,K

The equilibrium temperature of the system is 276.494 Kelvin.

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