because all organic compounds have in common the presence of carbon atoms and hydrogen atoms.
Option D, atomic number is the correct answer
Answer:
k = 6.31 x 10⁻³ min⁻¹
Explanation:
The equation required to solve this question is:
k = 0693 / t half-life
This equation is derived from the the equation from the radioctive first order reactions:
ln At/A₀ = -kt
where At is the number of isoopes after a time t , and A₀ is the number of of isotopes initially. The half-life is when the number of isotopes has decayed by a half, so
ln(1/2) = -kt half-life
-0.693 = - k t half-life
t half-life = 109.8 min
⇒ k = 0.693 / t half-life = 0.693 / 109.8 min = 6.31 x 10⁻³ min⁻¹
Answer:
- <em>The molar concentraion is </em><u>0.74 M</u>
Explanation:
<u>1) Data:</u>
a) % w/v = 2.5%
b) compound: H₂O₂ (from a table molar mass = 34.0147 g/mol)
c) M = ?
<u>2) Formulae:</u>
a) % w/v = (mass of soulte / volume of solution) × 100
b) numer of moles, n = mass in grams / molar mass
c) M = number of moles of solute / liters of solution
<u>3) Solution:</u>
a) T<u>ake a base of 100 ml of solution (0.100 liter)</u>:
- %w/v = 2.5% = 2.5 g solute / 100 ml solution
- mass of solute = 2.5 g / 100 ml × 100 ml = 2.5 g
b) <u>Calculate the number of moles of solute, n</u>:
- n = mass in grams / molar mass = 2.5 g / 34.0147 g/mol = 0.0735 mol
c) <u>Calculate the molarity, M</u>:
- M = n / V in liter = 0.0735 mol / 0.100 liter = 0.735 M
Round to two significant figures: 0.74 M ← answer
The application of acid-base chemistry is used to formulate an answer for your question.
moles NaOH = c · V = 0.2321 mmol/mL · 26.34 mL = 6.113514 mmol
moles H2SO4 = 6.113514 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.056757 mmol
Hence
[H2SO4]= n/V = 3.056757 mmol / 36.43 mL = 0.08391 M
The answer to this question is [H2SO4] = 0.08391 M