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erma4kov [3.2K]
3 years ago
12

When a atom gets smaller, what happens to the electronegativity ?

Chemistry
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

Increases electronegativity.

Explanation:

Hit that Thanks button, pls.

<Nathen>

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An iron bar at 200c is placed in thermal contact with an identical iron bar at 120c in an isolated system
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Conduction: In the conduction, the heat is transferred from the hotter body to the colder body until the temperature on both bodies are equal.

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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 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu
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Answer: The final temperature of nickel and water is  25.2^{o}C.

Explanation:

The given data is as follows.

   Mass of water, m = 55.0 g,

  Initial temp, (t_{i}) = 23^{o}C,      

  Final temp, (t_{f}) = ?,

  Specific heat of water = 4.184 J/g^{o}C,      

Now, we will calculate the heat energy as follows.

           q = mS \Delta t

              = 55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)

Also,

    mass of Ni, m = 15.0 g,

   Initial temperature, t_{i} = 100^{o}C,

   Final temperature, t_{f} = ?

 Specific heat of nickel = 0.444 J/g^{o}C

Hence, we will calculate the heat energy as follows.

          q = mS \Delta t

             = 15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)      

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

              q_{water}(gain) = -q_{alloy}(lost)

55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C) = -(15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C))

       t_{f} = \frac{25.9^{o}C}{1.029}

                 = 25.2^{o}C

Thus, we can conclude that the final temperature of nickel and water is  25.2^{o}C.

6 0
3 years ago
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