4.1 h = 14760 s
<span>t 1/2 = ln 2 / k </span>
<span>k = rate reaction = 4.97 x 10^-5 </span>
<span>ln 0.045 / 0.36 = - 4.97 x 10^-5 t </span>
<span>2.08 = 4.97 x 10^-5 t </span>
<span>t = 41839.9 s = 11 h 37 min 19 s</span>
Yes, the atomic radius increases as you move down a group of elements.
this is true
going down leads to valence electrons that are further away from nucleus -> less electrostatic attraction -> less pull towards nuc. -> greater radius/volume taken
₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴ + 2 ₀n¹
Mass of reactants = 235.04393 + 1.008665 = 236.052595 amu
Mass of products = 139.92144 + 93.91523 + 2* (1.008665) = 235.854000 amu
Mass defect Δ m = 236.052595 - 235.854000 = 0.198 amu
Reaction energy released Q = Δ m * 931.5
= 0.198 * 931.5 = 185 MeV
Answer:
T₂ = 169.89 K
Explanation:
Given data:
Initial volume = 250 cm³
Initial temperature = 10°C (10+273.15 K = 283.15 K)
Final temperature = ?
Final volume = 150 cm³
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = T₁V₂/V₁
T₂ = 283.15 K × 150 cm³ / 250 cm³
T₂ = 42472.5 K. cm³ / 250 cm³
T₂ = 169.89 K
Answer:
The answer to your question is -2855 J
Explanation:
Reaction
2C₂H₆ + 7O₂ ⇒ 4CO₂ + 6H₂O
Formula
Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants
Substitution
ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}
ΔHrxn = {-1574 -1450.8} - {-169.4}
ΔHrxn = -3024.8 + 169.4
ΔHrxn = -2855.4 J
