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
Single displacement and combustion reactions are ALWAYS redox.
Answer:
C= 0.532M
Explanation:
The equation of reaction is
H2SO4 + 2KOH = K2SO4+ H2O
nA= 1, nB= 2, CA= ?, VA= 48.9ml, CB= 1.5M, VB= 34.7ml
Applying
CAVA/CBVB = nA/nB
(CA× 48.9)/(1.5×34.7)= 1/2
Simplify
CA= 0.532M
Answer:- oxygen.
Explanations:- The electronic configuration is given and we are asked to figure out the electrically neutral atom that will have the electron configuration, .
The sum of electrons for this electron configuration is 8. If we look at the periodic table then 8 is the atomic number of oxygen.
So, the electrically neutral atom for the given electron configuration is oxygen.