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iVinArrow [24]
3 years ago
12

An iron bar at 200c is placed in thermal contact with an identical iron bar at 120c in an isolated system

Chemistry
2 answers:
Andrei [34K]3 years ago
8 0

Answer:

A.  

The temperatures of the iron bars after 30 minutes would be less than 160°C because heat would be lost to the surroundings.

Explanation:

For edmentum and plato

Step2247 [10]3 years ago
7 0

Conduction: In the conduction, the heat is transferred from the hotter body to the colder body until the temperature on both bodies are equal.

In thermal equilibrium, there is no heat transfer as the heat is transferred till the temperature on the bodies are not same.

In the given problem, an iron bar at 200°C is placed in thermal contact with an identical iron bar at 120°C in an isolated system. After 30 minutes, the thermal equilibrium is attained. Then, the temperature on both iron bars are equal.Both iron bars are at 160°C in an isolated system.

But in an open system, the temperatures of the iron bars after 30 minutes would be less than 160°C. There will be heat lost to the surrounding. The room temperature is 25°C. There will be exchange of the heat occur between the iron bars and the surrounding. But It would take more than 30 minutes for both iron bars to reach 160°C because heat would be transferred less efficiently.

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A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Cal
Nesterboy [21]

Answer:

Density is: 1.05 g/ml

Mole fraction solute: 0.015

Mole fraction solvent:  0.095

Molarity: 0.80 M

Molality: 0.82 m

Explanation:

A typical excersise of solution.

It is more confortable to make a table for this.

                |   masss  |  volume  |  mol

solute       |                |                |          

solvent     |                |                |  

solution    |                |                |

Let's complete, what we have.

                 |   masss  |  volume  |  mol

solute       |  10.8g     |                |          

solvent     |                |  133 mL   |  

solution    |                |  137 mL    |

We can first, know how many moles are 10.8 g

Molar Mass H3PO4 = 97.99 g/mol

Mass / Molar mass = mol

10.8 g / 97.99 g/m = 0.110 mol

Density of water is 1 g/ml (it is a very knowly value)

From this data, we can know water mass, solvent.

Density = mass / volume

1 g/ml = mass / 133 mL

Mass = 133 g

We can also have the moles, by the molar mass of water 18 g/m

133 g / 18 g/m = 7.39 mol

                 |   masss  |  volume  |  mol

solute       |   10.8g     |                |   0.110 mol      

solvent     |   133g      |  133 mL   |  7.39 mol

solution    |   143.8g   |  137 mL   | 7.50 mol

Mass of solution will be solute mass + solvent mass

Moles of solution will be solute moles + solvent moles

Now we can calculate everything.

Molarity means mol of solute in 1 L of solution. (mol/L)

We have to convert 137 mL in L (/1000)

0.137L so → 0.110 m / 0.137L = 0.80 M

Molality means mol of solute in 1kg of solvent.

We have to convert 133g in kg (/1000)

0.133 kg so → 0.110 m/0.133 kg = 0.82 m

Density is mass / volume

Solution density will be solution mass / solution volume

143.8 g/137 mL = 1.05 g/m

Molar fraction is : solute moles / total moles  or  solvent moles/total moles.

You can also (x 100%) to have a percent of them.

Remember sum of molar fraction = 1

Molar fraction of solute = 0.110 mol / 7.50mol = 0.015

Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985

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