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Write a balanced nuclear equation for the beta-emission of carbon-14 and calculate the remaining mass of a 3.20 g sample after 11460 years.

Chemistry

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(ii) Write a balanced nuclear equation for the β\beta-emission of carbon-14.
In β\beta-emission (beta decay), a neutron in the nucleus transforms into a proton, emitting an electron (−10e^0_{-1}\text{e}) and an antineutrino. The mass number remains the same, and the atomic number increases by 1.

614C⟶714N+−10e^{14}_{6}\text{C} \longrightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e}

The daughter nucleus is Nitrogen-14.

614C⟶714N+−10e\boxed{^{14}_{6}\text{C} \longrightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e}}

(iii) Calculate the mass that remains when a 3.20 g sample of carbon-14 undergoes decay for 11460 years.

  1. Identify the given values.

    Initial mass, N0=3.20 gN_0 = 3.20 \text{ g}
    Time elapsed, t=11460 yearst = 11460 \text{ years}
    Half-life of carbon-14, t1/2=5730 yearst_{1/2} = 5730 \text{ years} (from the context above the question)

  2. Calculate the number of half-lives (nn) that have occurred.

    n=tt1/2n = \frac{t}{t_{1/2}} n=11460 years5730 yearsn = \frac{11460 \text{ years}}{5730 \text{ years}} n=2n = 2

  3. Use the formula for remaining mass after nn half-lives.

    N(t)=N0(12)nN(t) = N_0 \left(\frac{1}{2}\right)^n N(t)=3.20 g×(12)2N(t) = 3.20 \text{ g} \times \left(\frac{1}{2}\right)^2 N(t)=3.20 g×14N(t) = 3.20 \text{ g} \times \frac{1}{4} N(t)=0.80 gN(t) = 0.80 \text{ g}

    The mass that remains is 0.80 g.

    0.80 g\boxed{0.80 \text{ g}}

    (c)(i) State the experimental method used in the determination of the molar mass of the liquid.
    The experimental method described, involving the vaporization of a known mass of a volatile liquid and measuring its volume, temperature, and pressure, is typically known as the Victor Meyer method or Dumas method. It relies on the Ideal Gas Law.

    Victor Meyer method (or Dumas method)\boxed{\text{Victor Meyer method (or Dumas method)}}

    (c)(ii) Calculate the molar mass of the liquid.

  4. Identify the given values and convert units where necessary.

    Mass of liquid, m=1.25 gm = 1.25 \text{ g}
    Temperature, T=75 °C=75+273.15=348.15 KT = 75 \text{ °C} = 75 + 273.15 = 348.15 \text{ K}
    Pressure, P=1.1 atmP = 1.1 \text{ atm}
    Volume of vapour, V=365 cm3=0.365 LV = 365 \text{ cm}^3 = 0.365 \text{ L}
    Ideal gas constant, R=0.08206 L atm mol−1 K−1R = 0.08206 \text{ L atm mol}^{-1} \text{ K}^{-1}

  5. Use the Ideal Gas Law, PV=nRTPV = nRT, where n=mMn = \frac{m}{M} (moles = mass/molar mass).

    Rearrange the formula to solve for molar mass (MM):

    PV=mMRTPV = \frac{m}{M}RT M=mRTPVM = \frac{mRT}{PV}

  6. Substitute the values into the formula and calculate MM.

    M=(1.25 g)(0.08206 L atm mol−1 K−1)(348.15 K)(1.1 atm)(0.365 L)M = \frac{(1.25 \text{ g})(0.08206 \text{ L atm mol}^{-1} \text{ K}^{-1})(348.15 \text{ K})}{(1.1 \text{ atm})(0.365 \text{ L})} M=35.78037875 g L atm mol−10.4015 L atmM = \frac{35.78037875 \text{ g L atm mol}^{-1}}{0.4015 \text{ L atm}} M≈89.12 g/molM \approx 89.12 \text{ g/mol}

    The molar mass of the liquid is approximately 89.1 g/mol.

    89.1 g/mol\boxed{89.1 \text{ g/mol}}
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