The energy transferred is 28.5 j
<em><u>calculation</u></em>
energy is calculate MCΔT formula where,
M(mass)= 6.30 grams
C(specific heat capacity)= 0.377 j/g
ΔT(change in temperature)= 32.0c- 20 c= 12 c
Energy is therefore= 6.30 g x 0.377 j/g /c x 12 c =<u>28.5 j</u>
<span>Ka is an equilibrium constant for the partial ionization of "weak" acids in water.</span>
The number of sodium ions is equal to the number of formula units of salt.
Answer:
The given element is Radon because its atomic weight is 222 amu.
Explanation:
Given data:
Percentage of A-219 = 13.92%
Percentage of B-222 = 72.16%
Percentage of C-225 = 13.92%
Atomic weight of element = ?
Solution:
Average atomic mass = (abundance of A isotope × its atomic mass) +(abundance of B isotope × its atomic mass) + (abundance of C isotope × its atomic mass) / 100
Average atomic mass = (13.92×219)+(72.16×222) + (13.92×225)/100
Average atomic mass = 3048.48 + 16019.52 +3132/ 100
Average atomic mass = 22200 / 100
Average atomic mass = 222 amu.
The given element is Radon because its atomic weight is 222 amu.