The answer is: K is more reactive than Ca because K has to lose only one electron to complete its outermost shell.
Potassium is a chemical element with atomic number 19 (number of electrons is 19).
Electron configuration of potassium is: ₁₉K 1s²2s²2p⁶3s²3p⁶4s¹.
Potassium is the alkali metal and has a single valence electron in the outer electron shell.
Periodic law is the arrangement of the elements in order of increasing atomic number.
For example all alkaline metals (I group of periodic table, Na, K, Cs...) loose one electron in chemical reaction and react vigorously with water.
Reactivity series is an empirical progression of a series of metals, arranged by their reactivity from highest to lowest (alkaline metals have highest reactivity and Noble metals lowest reactivity).
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons (one electron, earth alkaline metals (right next to alkaline metals) have higher ionization energy than alkaline metals, because they have two valence electrons.
Answer:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
Answer:
a) Sulphur + Oxygen → Sulphur dioxide
b) Carbon + Oxygen → Carbon dioxide
c) Sulphur + Iron → Iron sulphide