One atom of carbon weighs exactly 12/6.022x10^23 = 1.9927x10^-23 grams<span>.</span>
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: 300 K
Explanation:
Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.
(At constant pressure and number of moles)

Given : V= 6.0 L
k= 0.020 L/K
T=?


Thus temperature of the gas is 300 K.
1.33 is the answer thank me later
Al2(SO4)3 + 3Ca(OH)2 -> 2Al(OH)3 + 3Ca(SO)4