Radioactive decay => C = Co { e ^ (- kt) |
Data:
Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min
Time conversion: 4 hr 39 min = 4.65 hr
1) Replace the data in the equation to find k
C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}
=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719
2) Use C / Co = 1/2 to find the hallf-life
C / Co = e ^ (-kt) => -kt = ln (C / Co)
=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k
t = ln(2) / 0.44719 = 1.55 hr.
Answer: 1.55 hr
Answer:
Because it can cause health problems or injuries to our sense organs.
Explanation:
Chemicals in the laboratory are made up of different constituents, which may be dangerous or injurious to health. This is the reason why safety measures or precautions have to be taken when working in the laboratory. One of those safety measures is that "one should never use taste, touch, or smell to identify an unknown chemical".
This is so because a chemical that is unknown amounts to the fact that what such chemical contains is unknown, hence, the chemical might have the ability to cause harm or injuries to the sense organ. For example, a conc. acid that is tasted will burn the tongue etc.
Answer:
Cp = 0.093 J.g⁻¹.°C⁻¹
Solution:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 300 J
m = mass = 267 g
Cp = Specific Heat Capacity = ??
ΔT = Change in Temperature = 12 °C
Solving eq. 1 for Cp,
Cp = Q / m ΔT
Putting values,
Cp = 300 J / (267 g × 12 °C)
Cp = 0.093 J.g⁻¹.°C⁻¹
The epicardium (external layer), the myocardium (middle layer) and the endocardium (inner layer).
Answer:
531$
Explanation:
9 × 59 = 531 , 3 mo free, 12-3=9
