<u>Given:</u>
Initial amount of carbon, A₀ = 16 g
Decay model = 16exp(-0.000121t)
t = 90769076 years
<u>To determine:</u>
the amount of C-14 after 90769076 years
<u>Explanation:</u>
The radioactive decay model can be expressed as:
A = A₀exp(-kt)
where A = concentration of the radioactive species after time t
A₀ = initial concentration
k = decay constant
Based on the given data :
A = 16 * exp(-0.000121*90769076) = 16(0) = 0
Ans: Based on the decay model there will be no C-14 left after 90769076 years
C becuase that is the one that you would have to do becuase that is the only option
Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
We have to get the relationship between metallic character and atomic radius.
Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.
If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.
With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.
With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.
Grams of Phosphorus = 4.14 grams
Grams of white compound = 27.8 grams
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
Calculating moles which would be grams / molar mass
Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
Calculating the ratios by dividing with the small entity
P = 0.1337 moles / 0.1337 moles = 1
Cl = 0.6674 moles / 0.1337 moles = 5
So the empirical formula would be PCl5