Answer:
39.2 g
Explanation:
- 2Ni₂O₃(s) ⟶ 4Ni(s) + 3O₂(g)
First we <u>convert 55.3 grams of Ni₂O₃ into moles of Ni₂O₃</u>, using its<em> molar mass</em>:
- 55.3 g ÷ 165.39 g/mol = 0.334 mol Ni₂O₃
Then we <u>convert 0.334 moles of Ni₂O₃ into moles of Ni</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 0.334 mol Ni₂O₃ *
= 0.668 mol Ni
Finally we <u>calculate how much do 0.668 Ni moles weigh</u>, using the<em> molar mass of Ni </em>:
- 0.668 mol Ni * 58.69 g/mol = 39.2 g
Answer:
179 L of CO2
Explanation:
Given the equation of the reaction;
C2H6(g) + 7/2 O2(g) -------> 2CO2(g) + 3H2O(g)
Now 1 mole of ethane yields 2 moles of CO2 from the balanced reaction equation
1 mole of a gas occupies 22.4 L volume so,
22.4 L of ethane yields 44.8 L of CO2
89.5 L of ethane yields 89.5 * 44.8/22.4 = 179 L of CO2
Molar mass of C: 12.011 g/mol
The equation says C20, which means there are 20 carbon atoms in each molecule of Vitamin A. So, we multiply 12.011 by 20 to get 240.22 g/mol carbon.
Molar mass of H: 1.0079 g/mol
The equation says C30, which means there are 30 hydrogen atoms in each molecule of Vitamin A. So, we multiply 1.0079 by 30 to get 30.237 g/mol hydrogen.
Molar mass of O: 15.999 g/mol
The equation says O without a number, which means there is only one oxygen atom in each molecule of Vitamin A. So, we leave O at 15.999 g/mol.
Then, just add it up:
240.22 g/mol C + 30.237 g/mol H + 15.999 g/mol O = 286.456 g/mol C20H30O
So, the molar mass of Vitamin A, C20H30O, is approximately 286.5 g/mol.
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.
