Problem 2
You begin with 216 micrograms of Fermium - 253. After three days, the quantity halves, resulting in 108 micrograms left.
Another three days pass. Beginning with 108 micrograms, this amount gets halved again, leaving 54 micrograms.
Finally, after another three-day span, starting from 54 micrograms, you again halve this amount to reach 27 micrograms.
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
Your example is Nitrogen. Begin by completing the table, then formulate some rules to help prepare for possible alternate elements in the test. This approach is quite useful.
Table
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
Rules
As the number of bonds INCREASES, the energy within the bond also INCREASES
As the number of bonds INCREASES, the distance of the bond DECREASES.
Answer:
2(CH3)2N2H2 + 3N2O4 → 4N2 + 4H2O + 4CO2 + heat
Explanation:
- To balance chemical equations, coefficients are assigned to both reactants and products.
- This yields an equal count of atoms of each element on both sides of the equation.
- Balancing chemical equations ensures compliance with the law of conservation of mass.
- According to this law, the mass of reactants must equal the mass of products, achievable through balancing the equation.
- The application of coefficients 2, 3, 4, 4, 4 allows for an equal balance in the equation.
- Consequently, the balanced equation can be written as:
2(CH3)2N2H2 + 3N2O4 → 4N2 + 4H2O + 4CO2 + heat
<span>According to crystal field strength, the Cl ligand results in the longest d-d transition when coordinated with Ti(III) due to its classification as a weak field ligand that causes minimal d orbital splitting.</span>
When two atoms with equal electronegativity bond together, they form nonpolar covalent bonds.
Your second statement mirrors the first; the second statement likely reads, "Bonds between two atoms with unequal electronegativity are termed polar covalent bonds."
N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).
