IB Chemistry Make your chemistry assessable worked solutions

Count from the first non-zero digit: 4, the interior zero, 7 and the recorded final zero. There are four significant figures, so choose C. Leading zeros locate the decimal point.

Rearrange the concentration approximation to obtain [\mathrm{H^+}]=10^{-\mathrm{pH}}\,\mathrm{mol\,dm^{-3}}. Substitute 5.28 to get 5.2480746\times10^{-6}\,\mathrm{mol\,dm^{-3}}. A is the closest option to this unrounded value. For a measured concentration report, two decimal places in pH imply two significant figures: 5.2\times10^{-6}\,\mathrm{mol\,dm^{-3}}.

The tested interval ends at 14.0 mm, whereas the unknown corresponds to 21.0 mm. Choose C: standards near the unknown test the relationship where it is being used. Repeats at 9.0 mm only strengthen evidence inside the original range.

(a) Count 5+2(1)=7 valence electrons, then subtract the four bonding electrons to leave 7-4=3 [1]. (b) Retain both N-H bonds. Place one lone pair [1] and one unpaired electron [1] on N. The neutral fragment is a radical. The separate original Lewis graphic shows the electron pair above N and the single electron below.

Breaking the C-C and C-H bonds requires substantial energy because these bonds are strong [1]. Their small polarity provides little charge separation for attack by polar reagents [1]. The proposed weak-bond explanation cannot account for low reactivity.

Choose B. The given coefficient describes how closely the straight-line model accounts for variation in the measured response. It cannot establish the gradient sign or a causal mechanism. It also supplies no per-reading uncertainty.

(a) Substitute into [\mathrm{H^+}]=10^{-\mathrm{pH}}\,\mathrm{mol\,dm^{-3}} to obtain 10^{-4.32}=4.7863009\times10^{-5}. Report 4.8\times10^{-5}\,\mathrm{mol\,dm^{-3}} [1]. (b) Use \mathrm{pH}=-\log_{10}([\mathrm{H^+}]/(\mathrm{mol\,dm^{-3}})). This gives -\log_{10}(7.9\times10^{-3})=2.1023729, reported as 2.10 [1]. (c) Calibration connects the probe signal to numerical reference values [1]. Indicator colour matching spans a range and cannot justify the requested decimal resolution [1]. Resolution alone does not establish accuracy.

(a) Calculate the arithmetic mean: \bar m=60.48/4=15.12\,\mathrm{g} [1]. Subtract the extreme readings to obtain a range of 15.13-15.11=0.02\,\mathrm{g} [1]. (b) The tight grouping supports good precision [1]. Every reading exceeds the exact reference mass; the mean bias is 15.12-15.00=+0.12\,\mathrm{g} [1]. Repeating a measurement does not remove this common offset.

(a) Calculate 1/t to obtain 0.00833, 0.0167 and 0.0250\,\mathrm{s^{-1}} [1]. The reciprocal times change by the same factors as concentration, supporting direct proportionality of this proxy over the tested range [1]. (b) Each trial reaches the same small product concentration, so its mean formation rate has a fixed concentration increment in the numerator [1]. Negligible reactant depletion makes that mean rate a useful starting-rate approximation [1]. (c) Repeat each setting and add concentrations, then compare the relationship with the scatter [1]. The proxy itself has inverse-time units; it is not a concentration-per-time rate or a mechanism determination.

Use the supplied M=142.05\,\mathrm{g\,mol^{-1}}. (a) Convert the stock volume to 0.1000\,\mathrm{dm^3} and calculate n=cV=0.1200\times0.1000=0.01200\,\mathrm{mol} [1]. Then m=nM=1.7046\,\mathrm{g}, reported as 1.705 g [1]. (b) Transfer the aliquot using a 10.00\,\mathrm{cm^3} volumetric pipette [1]. Conserved solute gives c_2=c_1V_1/V_2=0.1200\times10.00/250.0=0.004800\,\mathrm{mol\,dm^{-3}} [1]. (c) The sample contains 0.004800\times0.05000=2.400\times10^{-4}\,\mathrm{mol} sulfate. The supplied 1:1 ion equation requires the same amount of barium ions [1]. (d) Overfilling increases volume at unchanged solute amount, so the concentration is too low [1].

(a) Weigh the salt and water independently [1]. Establish the water's starting temperature [1]. Add the salt, cover and mix until dissolution is complete [1]. Record a representative endpoint temperature and justify any temperature-time extrapolation used to estimate the change [1]. (b) Use the complete solution mass, m_{\mathrm{solution}}=m_{\mathrm{water}}+m_{\mathrm{salt}}, and calculate q_{\mathrm{solution}}=m_{\mathrm{solution}}c_p\Delta T, where \Delta T=T_{\mathrm{final}}-T_{\mathrm{initial}} [1]. Calculate n=m_{\mathrm{salt}}/M and then \Delta H/(\mathrm{kJ\,mol^{-1}})=-q_{\mathrm{solution}}/(1000n) for heat in joules [1]. Here c_p is in \mathrm{J\,g^{-1}\,K^{-1}} and masses are in grams. Cooling gives positive dissolution enthalpy. This uses the assumed known solution heat capacity and negligible cup heat capacity; it does not silently assume that external heat exchange is absent in an actual experiment.

(a) Label concentration and rate axes with their units [1]. Plot (0.040,0.0016), (0.080,0.0032) and (0.120,0.0048), and draw their straight line through the origin [1]. (b) Rate scales directly with concentration, indicating first order and r=k[\mathrm Q] [1]. (c) Rearrange to obtain k=r/[\mathrm Q]=0.0016/0.040=0.040 [1]. Concentration units cancel, leaving \mathrm{s^{-1}} [1]. (d) Predict r=k[\mathrm Q]=0.040\times0.100=4.0\times10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}} [1]. The separate original answer graph uses these three coordinates and the origin.