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).
1. The total moles of the solution is 0.3079193 mol.
2. The mole fraction for gold is 0.2473212, and for silver, it is 0.7526787.
3. The molar entropy of mixing for gold is 2.87285 j/K, while for silver, it is 1.77804 j/K.
4. The total entropy of mixing sums to 4.65089 j/K.
5. Molar free energy amounts to -2325.445 kJ.
6. Chemical potential for silver is -1750.31129 j/mol and for gold, it is -575.13185 j/mol. To elaborate:
(1) The molar mass of silver stands at 107.8682 g/mol, and gold's at 196.96657 g/mol. Hence, calculating moles leads to mass/molar mass for silver: 25 g/107.8682 g/mol = 0.2317643 mol and for gold: 15 g/196.96657 g/mol = 0.076155 mol, resulting in a total of 0.30791193 mol.
(2) For the mole fractions, silver's fraction is 0.2317643/0.3079193 = 0.7526787, and for gold, it's 0.076155/0.3079193 = 0.2473212.
(3) To find molar entropy mixing ∆Sm, we use the formula ∆Sm = -R * Xi * ln(Xi) where R = 8.3144598. For silver, substituting gives us 1.77804 j/K, while for gold, we get 2.87285 j/K.
(4) Overall entropy of mixing totals 4.65089 j/K thus calculated.
(5) The Gibbs free energy at 500 °C can be derived through G = H - TS, accounting to H = 0 (as T is 500 + 273 = 773 K and S is 4.65089), resulting in G equating to -3595.138 kJ.
(6) The chemical potentials calculated derive from multiplying the Gibbs free energy by their mole fractions.
Answer:
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Explanation:
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<span>The mass percentage represents the weight of a substance within a mixture in relation to a total mass of 100 units. It's also referred to as %w/w. To find this, you divide the mass of the solute by the total mass of the mixture. Given that we know the volume of water, we need to convert that volume into mass using its density. The calculation is as follows:
Mass of the solution = 100 mL (0.99993 g/mL) water + 25 g EtOH
Mass of the solution = 124.993 g of solution
%w/w = (25 g / 124.993 g) x 100
%w/w = 20% for EtOH</span>
Answer:
The accurate assertion is "Sample 1 will have a higher percent transmittance of light, %T, than Sample 2."
Explanation:
The Beer-Lambert Law indicates that absorbance correlates directly with the concentration of a solution.
Absorbance = εLc
ε: molar absorptivity coefficient
L: length of the light path
c: concentration
Thus, assuming ε and L remain constant, an increase in concentration results in greater absorbance and reduced transmittance.
Given that sample 1 has a greater CuSO₄ concentration than sample 2, it subsequently exhibits higher absorbance and lower percent transmittance.
