Answer:
Kc = 0.5951 (4 sig. figs.)
Explanation:
For A + B ⇄ C + D at standard thermodynamic conditions (298K, 1atm)
ΔG = ΔG° + R·T·lnQ => 0 = ΔG° + R·T·lnKc => ΔG° = - R·T·lnKc
=> lnKc = - ΔG°/R·T
ΔG° = +12.86 Kj/mol
R = 8.314 Kj/mol·K
T = 298K
lnKc = - (+12.86Kj) / (8.314Kj/mol·K)(298K) = - 0.519 mol⁻¹
Kc = e⁻⁰°⁵¹⁹ mol⁻¹ = 0.5957 mol⁻¹ (4 sig. figs.)
Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 3(NH4)2SO4(aq) + Cr2(CO3)3(s)
<span>Ionic: 2Cr+3 + 3SO4^-2 + 6NH4+ + 3CO3^-2 ----> 6NH4+ + 3SO4^-2 + Cr2(CO3)3 (spectator ions are NH4+, SO4^-2) </span>
<span>Net Ionic: 2Cr^+3(aq) + 3CO3^-2(aq) -------> Cr2(CO3)3(s) </span>
Answer:
B. It is important that people are not harmed for the sake of science.
Explanation:
Ethical principles stress the need to do good and cause no harm.A researcher is therefore required to;
- obtain an informed consent from the participants
- minimize or eliminate risk of harm to participants
- protect the anonymity and confidentiality of participants
- Apply no deceptive techniques
- allow the right to withdraw from the study by a participant
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.
Solution:
At the equivalence point, moles NaOH = moles benzoic acid
HA + NaOH ==> NaA + H2O where HA is benzoic acid
At the equivalence point, all the benzoic acid ==> sodium benzoate
A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)
Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11
Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150
x^2 = 3.33x10^-12
x = 1.8x10^-6 = [OH-]
pOH = -log [OH-] = 5.74
pH = 14 - pOH = 8.26
