Atomic mass Si = 28.0855 u.m.a
28.0855 g -------------------- 6.02x10²³ atoms
3.96x10³ g ------------------- ?? atoms
( 3.96x10³) x 6.02x10²³ / 28.0855 =
8.48x10²⁵ atoms
An original sample of K-40 has a mass of 25.00 grams.
After 3.9 m 109 years, 3.125 grams of the original sample remains unchanged.
What is the half-life of K-40?
First step is to determine the remaining decimal amount.
3.125 grams /25.00 grams = 0.125
Second step is to determine the number of half lives.
(1/2)^n = 0.125
N log (1/2) = log 0.125
N = 3 years
4 significant figures. Use a rule called the Atlantic-Pacific rule where if the period is absent (Atlantic), you would start counting the numbers from right to left. If the period is present (Pacific), then start counting the numbers from left to right. Since the period is present, start from 5 and there are 4 digits including 5.
Answer:
Explanation:
<u>1) Data:</u>
a) n = ?
b) V = 50.0 dm³ = 50.0 liter
c) p = 100.0 kPa
d) T = 50 °C
<u>2) Physical law:</u>
<u>3) Constants:</u>
- R = 0.08206 atm-liter / K-mol
<u>4) Unit conversions:</u>
- T = 50 + 273.15 K = 323.15 K
- P = 100.0 kPa × (1 amt /101,325 kPa) = 0.9869 atm
<u>5) Solution:</u>
- n = 0.9869 atm × 50.0 liter / (0.08206 atm-liter /K-mol × 323.15 K)
Mass of the bullet (m) = 8.13 g = 0.00813 kg
velocity of the bullet (v) = 1337 miles/hr
The de Broglie wavelength (∧) of an object of mass (m) travelling at a velocity (v) is given as:
∧ = h/m*v
where h = Planck's constant = 6.626 *10^-34 kgm2/s
Convert velocity from miles/hr to m/s
1 mile/hr = 0.447 m/s
1337 miles/hr = 0.447*1337 = 597.64 m/s
∧ = 6.626*10^-34 m2.Kg.s-1/0.00813 kg * 597.64 ms-1
∧ = 1.363*10^-34 m
