Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value
<span>
The half-life for the chemical reaction is 29,2 s and is
independent of initial concentration.
c</span>₀
- initial concentration the reactant.
c - concentration of the reactant remaining
at time.
t = 29,2 s.<span>
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s</span> = 0,0237 1/s.<span>
ln(c/c</span>₀) = -k·t₁.<span>
ln(1/16 </span>÷ 1) = -0,0237 1/s ·
t₁.
t₁ = 116,8 s.
Answer:
The answer to your question is the letter C) 5648 kJ/mol
Explanation:
Data
C₁₂H₂₂O₁₁ + 12 O₂ ⇒ 12 CO₂ + 11 H₂O
H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol
H° O₂ = 0 kJ / mol
H° CO₂ = -393.5 kJ/mol
H° H₂O = -285.8 kJ/mol
Formula
ΔH° = ∑H° products - ∑H° reactants
Substitution
ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)
ΔH° = -4722 - 3143.8 + 2221.8
Result
ΔH° = -5644 kJ/mol
Answer:
с
Explanation:
the first quantum number of an electron gives the information about the energy level the electron is in
Answer:
1.25 gram of cesium-137 will remain.
Explanation:
Given data:
Half life of cesium-137 = 30 year
Mass of cesium-137 = 5.0 g
Mass remain after 60 years = ?
Solution:
Number of half lives passed = Time elapsed / half life
Number of half lives passed = 60 year / 30 year
Number of half lives passed = 2
At time zero = 5.0 g
At first half life = 5.0 g/2 = 2.5 g
At 2nd half life = 2.5 g/ 2 = 1.25 g
Thus. 1.25 gram of cesium-137 will remain.
