Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Answer:
I don't understand what you are asking
The answer to problem is [He] 2s1. Hope it helps
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