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Nutka1998 [239]
2 years ago
7

40.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 20.0°C in a coffee cup calorimeter. Calculate the final

temperature reached, assuming no heat loss or gain from the surroundings. (Data: specific heat capacity of H2O(l), c = 4.18 J/g×°C; H2O(s) => H2O(l) DH = 6.02 kJ/mol)Calculate the final temperature reached, assuming no heat loss or gain from the surroundings. (Data: specific heat capacity of H2O(l), c = 4.18 J/g×°C; H2O(s) => H2O(l) DH = 6.02 kJ/mol)
Chemistry
1 answer:
Stells [14]2 years ago
8 0

The final temperature of the mixture in the coffee cup calorimeter is; 19.467 °C

According to the law of energy conservation:

As such; the heat transfer in the liquid water is equal to heat gained by the ice

Heat transfer by liquid water is therefore;

  • <em>DH = m × c × DT</em>

  • DH = 6.02 kJ/mol) = 150 × 4.18 × (T1 - T2)

  • 6020 J/mol = 627 × (20 - T2)

However, since 18g of water makes one mole

  • 6020 J/mol = 6020/18 = 334.44 J/g.

  • 334.44 = 627 × (20 - T2)

  • 0.533 = (20 - T2)

  • T2 = 20 - 0.533

T2 = 19.467°C

Read more on heat:

brainly.com/question/21406849

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One hundred grams of water is saturated with NH CI at
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Answer:

  • About 18 g of NH₄Cl will precipitate.

Explanation:

The <em>table G</em> is the graph of the solubility curves for several solutes which is attached.

The second picture identifies the solubilities for the  NH₄Cl at 50ºC and 10ºC.

The solubility of NH₄Cl at 50ºC is about 52 g/ 100 g of water.

The solubility of NH₄Cl at 10ºC is about 34 g / 100 g of water.

Then, at 50ºC 100 g of water saturated with NH₄Cl contains about 52 g of NH₄Cl and 100 g of water saturated with NH₄Cl contains 34 g of NH₄Cl.

The difference, 52g - 34 g of NH₄Cl shall precipitate:

          52 g - 34 g = 18g ← answer

7 0
3 years ago
Round the following number to four figures: 45.4673
zhenek [66]

This says four figures. The 4 figures you should use are 4546  with a peek at 7 to see what effect it will have on the 4 main figures.

45.467 rounds to 45.47

You round the 4th figure up one. You are not concerned about the decimal places in this question.

7 0
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In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t
solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696  = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 4.81 moles

4 0
3 years ago
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Answer:

A B and C

Explanation:

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