With your mind. Boom. Lol I'm sorry I don't know I just need points
Answer:
6.2 calories
Explanation:
Data Given:
change in temperature = 20 °C
specific heat of gold = 0.031 calories/gram °C
mass of gold = 10.0 grams
Amount of Heat = ?
Solution:
Formula used
Q = Cs.m.ΔT
Where:
Q = amount of heat
Cs = specific heat of gold = 0.031 calories/gram °C
m = mass
ΔT = Change in temperature
Put values in above equation
Q = 0.031 calories/gram °C x 10.0 g x 20 °C
Q = 6.2 calories
So option A is correct = 6.2 calories
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.
