Answer:
100 g
Explanation:
From the question given above, the following data were obtained:
Original amount (N₀) = 400 g
Time (t) = 4 years
Half-life (t½) = 2 years
Amount remaining (N) =?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Time (t) = 4 years
Half-life (t½) = 2 years
Number of half-lives (n) =?
n = t / t½
n = 4 / 2
n = 2
Thus, 2 half-lives has elapsed.
Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:
Original amount (N₀) = 400 g
Number of half-lives (n) = 2
Amount remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2² × 400
N = 1/4 × 400
N = 0.25 × 400
N = 100 g
Thus, the amount of the radioactive isotope remaing is the 100 g.
Answer : The percent by mass of the solution is 7.81 %
Explanation : Given,
Mass of NaCl = 8.75 g
Mass of solution = 112.0 g
Now we have to determine the percent by mass of the solution.
Putting values in above equation, we get:
Therefore, the percent by mass of the solution is 7.81 %
Answer: An electron will jump to a higher energy level when excited by an external energy gain such as a large heat increase or the presence of an electrical field, or collision with another electron.
Explanation:
Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
