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)
The answer to the question is C
What is the average velocity of atoms in 1.00 mol of argon (a monatomic gas) at 275 k for m, use 0.0399kg
Answer: The average velocity of the atoms 847.33 m/s.
Explanation:
Moles of the neon = 1.00
Temperature of the gas : 288 K
Mass of the gas = 0.01000
R = 8.31 J/mol K
The average velocity of the atoms 847.33 m/s.
