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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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In the following reaction, which element or compound is the oxidizing agent?
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Vladimir79 [104]

First, we have to remember the molarity formula:

M=\text{ }\frac{moles\text{ of solute}}{L\text{ solution}}

Part 1:

In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:

M.W_{NaNO_3}=\text{ 23+14+16*3= 85 g/mol}

Then, we calculate the moles present in the solution:

3.976\text{ g NaNO}_3\text{ * }\frac{1\text{ mol}}{85\text{ g}}=\text{ 0.04678 mol NaNO}_3

Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):

M=\frac{0.04678\text{ mol}}{200\text{ mL*}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.2339 M}

The molarity of this solution equals 0.2339 M.

Part 2:

In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:

M=\text{ }\frac{0.04678\text{ mol}}{275\text{ mL *}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.1701 M}

So, the molarity of this solution is 0.1701 M.

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