F=1/T
therefore T=1/F
so thats T=1/20x10^6
T=5x10^-8seconds
Answer:
Calculate the mass of 6.022 × 1023 molecule of Calcium carbonate (CaCO3).
Solution —
Molar mass (Molecular mass in gram) of CaCO3 = 40+12+3×16 = 100 g
No. of moles of CaCO3
= No. of molecules/Avogadro constant
= 6.022 × 1023/ 6.022 × 1023
= 1 mole
Mass of CaCO3
= No. of moles × molar mass
= 1 × 100 g = 100 g.
It gets heavier and heavier
Answer:
v_average = 500 m / min
Explanation:
Average speed is defined
v = (x_{f} -x₀) / Δt
let's look in each section
section 1
the variation of the distance is 800 in a time of 1.4 min
v₁ = 800 / 1.4
v₁ = 571.4 m / min
section 2
distance interval 500 in a 1.6 min time interval
v₂ = 500 / 1.6
v₂ = 312.5 m / min
section 3
distance interval 1200 m in a time 2 min
v₃ = 1200/2
v₃ = 600 m / min
taking the speed of each section we can calculate the average speed
the distance traveled
Δx = 800 + 500 + 1200
Δx = 2500 m
the time spent
Δt = 1.4 + 1.6+ 2
Δt = 5 min
v_average = Δx / Δt
v_average = 2500/5
v_average = 500 m / min
That's true. The give-away is "observation ... in the real world". That's what "field reports" are.
The part I'm not so sure about is those "entific findings".
