Answer:
Conclusion
Explanation:
I believe you were asking for the term that best matches with the description given. Typically the conclusion summarizes your experiment in a 1 to 2 paragraph format.
We need (i) the stoichiometric equation, and (ii) the equivalent mass of dihydrogen.
Explanation:
1
2
N
2
(
g
)
+
3
2
H
2
(
g
)
→
N
H
3
(
g
)
11.27
g
of ammonia represents
11.27
⋅
g
17.03
⋅
g
⋅
m
o
l
−
1
=
?
?
m
o
l
.
Whatever this molar quantity is, it is clear from the stoichiometry of the reaction that 3/2 equiv of dihydrogen gas were required. How much dinitrogen gas was required?
Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).
Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.
Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
