Answer:
Oxide of M is and sulfate of
Explanation:
0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.
Let moles of hydrogen gas be n.
Temperature of the gas ,T= 17°C =290 K
Pressure of the gas ,P= 741 mmHg= 0.9633 atm
Volume occupied by gas , V = 0.303 L
Using an ideal gas equation:
Moles of hydrogen gas produced = 0.01225 mol
Moles of metal =
So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.
x = 2.9 ≈ 3
Formulas for the oxide and sulfate of M will be:
Oxide of M is and sulfate of .
2.4(10^3)
=2.4*10^3
=2.4*(10*10*10)
=2400 <span>milliliters
To </span>centiliters is<span> 2400mL= <u>240.0000cl. </u> </span>
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.
Answer:
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