We can calculate years by using the half-life equation. It is expressed as:
A = Ao e^-kt
<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span>From the half-life data, we can calculate for k.
1/2(Ao) = Ao e^-k(1620)
<span>k = 4.28 x 10^-4
</span>
0.125 = 1 e^-<span>4.28 x 10^-4 (</span>t)
t = 4259 years
<span>In the field of science, usually, the product of an experiment is
computed ahead to understand if it reached a specific objective. It would reach
greater than 100% of percent yield if the factors include faster reaction rates;
proper handling of the reactants, no outside contaminants, and the procedure of
the experiment is followed smoothly. It would reach lesser than 100% percent yield
if the experiment is not followed, external factors such as contamination from
the environment (wind, moisture, etc). </span>
Answer:
100 degree celcius, because it is the melting point of ice ob
Answer:
1.) 13 g C₄H₁₀
2.) 41 g CO₂
Explanation:
To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)
Molar Mass (CO₂): 44.007 g/mol
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O
48 g O₂ 1 mole 2 moles C₄H₁₀ 58.124 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 13 g C₄H₁₀
48 g O₂ 1 mole 8 moles CO₂ 44.007 g
--------------- x ----------------- x -------------------------- x ------------------ =
31.996 g 13 moles O₂ 1 mole
= 41 g CO₂
