Answer:
1. Find the molar mass of all the elements in the compound in grams per mole.
2. Find the molecular mass of the entire compound.
3. Divide the component's molar mass by the entire molecular mass.
4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.
Explanation:
Answer:
76.03 °C.
Explanation:
Equation:
C2H5OH(l) --> C2H5OH(g)
ΔHvaporization = ΔH(products) - ΔH (reactants)
= (-235.1 kJ/mol) - (-277.7 kK/mol)
= 42.6 kJ/mol.
ΔSvaporization = ΔS(products) - ΔS(reactants)
= 282.6 J/K.mol - 160.6 J/K.mol
= 122 J/K.mol
= 0.122 kJ/K.mol
Using gibbs free energy equation,
ΔG = ΔH - TΔS
ΔG = 0,
T = ΔH/ΔS
T = 42.6/0.122
= 349.18 K.
Coverting Kelvin to °C,
= 349.18 - 273.15
= 76.03 °C.
Answer:
For every pound lost, replace it with 16 to 20 ounces of fluid
Answer:
3,964 years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of the element is 5,730 years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 100%).
[A] is the remaining concentration of the sample ([A] = 61.9%).
∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.
