The number of Ml of C₅H₈ that can be made from 366 ml C₅H₁₂ is 314.7 ml of C₅H₈
<u><em>calculation</em></u>
step 1: write the equation for formation of C₅H₈
C₅H₁₂ → C₅H₈ + 2 H₂
Step 2: find the mass of C₅H₁₂
mass = density × volume
= 0.620 g/ml × 366 ml =226.92 g
Step 3: find moles Of C₅H₁₂
moles = mass÷ molar mass
from periodic table the molar mass of C₅H₁₂ = (12 x5) +( 1 x12) = 72 g/mol
moles = 226.92 g÷ 72 g/mol =3.152 moles
Step 4: use the mole ratio to determine the moles of C₅H₈
C₅H₁₂:C₅H₈ is 1:1 from equation above
Therefore the moles of C₅H₈ is also = 3.152 moles
Step 5: find the mass of C₅H₈
mass = moles x molar mass
from periodic table the molar mass of C₅H₈ = (12 x5) +( 1 x8) = 68 g/mol
= 3.152 moles x 68 g/mol = 214.34 g
Step 6: find Ml of C₅H₈
=mass / density
= 214.34 g/0.681 g/ml = 314.7 ml
I need more information I don’t understand the question
Answer:
Tryptophan is an essential amino acid.
Explanation:
Structural Formulas v. Empirical Formulas
An empirical formula (like a molecular formula) lacks any structural information about the positioning or bonding of atoms in a molecule. It can therefore describe a number of different structures, or isomers, with varying physical properties. For butane and isobutane, the empirical formula for both molecules is C2H5, and they share the same molecular formula, C4H10. However, one structural representation for butane is CH3CH2CH2CH3, while isobutane can be described using the structural formula (CH3)3CH.
Answer:
Case I => %C-14 remaining = 77% => Age of artifact = 2200 yrs
Case II => %C-14 remaining = 6.2% => Age of artifact = 23,000 yrs
Explanation:
Given:
Half-Life C-14 = 5730 yrs
=> Rate Constant = k = 0.693/t(1/2) = (0.693/5730)yrs⁻¹ = 1.2 x 10⁻⁴ yrs⁻¹
NOTE => All radioactive decay is 1st order kinetics.
=> A = A₀eˉᵏᵗ (classic 1st order decay equation)
- A = remaining activity
-A₀ = initial activity
- k = rate constant
- t = time of decay (or, age of object of interest; i.e., not everything is organic but the 1st order decay equation is good for non-organic objects (rocks) also. Analysts just use a different decay standard => K-40 → Ar-40 + β).
Solving the decay equation for time (t) ...
t = ln(A/A₀)/-k
Applying to problem cases...
Case I => %C-14 remaining = 77%
t = ln(A/A₀)/-k = ln(77/100)/-1.2x10⁻⁴ years = 2178 yrs ~ 2200 yrs
Case II => %C-14 remaining = 6.2%
t = ln(A/A₀)/-k = ln(6.2/100)/-1.2x10⁻⁴ years = 23,172 yrs ~ 23,000 yrs