Answer:
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Answer:
b) C = 0.50 J/(g°C)
Explanation:
∴ Q = 50 J
∴ m = 10.0 g
∴ ΔT = 35 - 25 = 10 °C
specific heat (C) :
⇒ C = Q / mΔT
⇒ C = 50 J / (10.0 g)(10 °C)
⇒ C = 0.50 J/(g°C)
Answer: Edge length of the unit cell = 628pm
Explanation: For a body centred cubic structured system, the relationship between the edge length of the unit cell and radius of the atoms in the structure is
Edge length of Unit cell (a) = (4R)/(√3)
R = 272pm = (272 × (10^-12))m = (2.72 × (10^-10))m
a = (4 × (2.72 × (10^-10)))/(√3)
a = (6.28157 × (10^-10))m = 628pm
When calcium carbonate is heated, it breaks down to form calcium oxide and carbon dioxide.
Thermal decomposition is the process in which heat is required.
It is also known as thermolysis.
It is processed in which a compound breaks into two or more products when the heat is supplied.
This reaction is used for the production of oxygen.
This reaction is also used for production of acidic as well as basic oxides.
CaCO3 on thermal decomposition gives:
CaCO3→CaO+CO2
CaO→ Basic oxide.
CO2→ Acidic oxide.
Answer:
23.2 g of Al will be left over when the reaction is complete
Explanation:
2Al + 3S → Al₂S₃
1 mol of Al = 26.98 g
1 mol of S = 32.06 g
Mole = Mass / Molar mass
63.8 g/ 26.98 g/m = 2.36 mole of Al
72.3 g / 32.06 g/m = 2.25 mole of S
2 mole of Aluminun react with 3 mole of sulfur
2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S
As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.
3 mole of S react with 2 mole of Al
2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole
I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.
2.36 mole of Al - 1.50 mole of Al = 0.86 mole
This is the quantity of Al without reaction.
Molar mass . mole = Mass → 26.98 g/m . 0.86 m = 23.2 g
