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Likurg_2 [28]
4 years ago
14

If 40.0 g of water at 70.0°c is mixed with 40.0 g of ethanol at 10.0°c, what is the final temperature of the mixture?

Chemistry
1 answer:
FinnZ [79.3K]4 years ago
3 0
When the heat lost by water = heat gained by ethanol

∴(  M* C * ΔT )w  = (M*C*ΔT ) eth

when Mw mass of water = 40 g 

C specific heat of water = 4.18

ΔT = (70- Tf) 

and M(eth) mass (ethanol) = 40 g 

C specific heat of ethanol = 2.44 

ΔT = (Tf - 10 ) 

by substitution:

40* 4.18 * (70 - Tf) = 40 * 2.44 * (Tf-10) 

∴Tf = 47.9 °C 
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A.) 4.0

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6 0
2 years ago
1) If 0.193 grams of toluene is dissolved in 2.532 grams of p-xylene, what is the molality of toluene in the solution?2) If a fr
AveGali [126]

Answer:

The value of K_f for xylene is 4.309°C/m.

The molar mass of pentane using this data is 73.82 g/mol.

Explanation:

\Delta T_f=K_f\times \frac{\text{Amount of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent(kg)}}

where,

\Delta T_f =depression in freezing point

K_f = freezing point constant  

we have :

1) freezing point constant  for xylene = K_f =?

Mass of toluene = 0.193 g

Mass of xylene = 2.532 kg = 0.002532 kg ( 1 g =0.001 kg)

\Delta T_f=3.57^oC

3.57^oC=K_f\times \frac{0.193 g}{92 g/mol\times 0.002532 kg}

K_f=4.309^oC/m

The value of K_f for xylene is 4.309°C/m.

2)

Mass of pentane = 0.123 g

molar mass of pentame= M

Mass of xylene = 2.493 g =  0.002493 kg

Freezing point Constant of xylene = K_f=4.309^oC/m

2.88^oC=4.309^oC/m\times \frac{0.123g}{M\times 0.002493 kg}

M = 73.82 g/mol

The molar mass of pentane using this data is 73.82 g/mol.

5 0
3 years ago
A hot piece of copper was dropped into 155 g of water at 23.6 °C and 2,100 J of energy was transferred to the water. What is the
BlackZzzverrR [31]

Answer:

26.8 °C

Explanation:

Step 1: Given and required data

  • Mass of water (m): 155 g
  • Energy transferred to the water (Q): 2100 J
  • Initial temperature of the water (T₂): 23.6 °C
  • Final temperature of the water (T₁): ?
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Step 2: Calculate the final temperature of the water

We will use the following expression.

Q = c × m × (T₂ - T₁)

T₂ = 2100 J/(4.184 J/g.°C) × 155 g + 23.6 °C = 26.8 °C

5 0
3 years ago
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