Using the G° data in your Appendix B, calculate the change in Gibbs free energy for each of the following reactions. In each case, indicate whether the reaction is spontaneous under standard conditions. a) H2 (g) + Cl2 (g) → 2HCl (g) b) MgCl2 (s) + H2O (l) → MgO (s) + 2 HCl (g) c) 2 NH3 (g) → N2H4 (g) + H2 (g) d) 2 NOCl (g) → 2 NO (g) + Cl2 (g) 4. From the values given for ΔH° and ΔS°, calculate ΔG° at 25°C for each of the following reactions. If the reaction is not spontaneous under standard conditions at 298K, at what temperature (if any) would the reaction become spontaneous?
100. g Al reacts with excess O2 to produce 150. g Al2O3 according to Calculate the theoretical and percentage yield. 4Al + 302 → 2 Al2O3. 79.4 % 14. Calculate the energy produced by the complete reaction of 150. g H2. 2H2 + O2 → 2H2O + 130KJ 4.83 x 103 kJ 15.
The distance between the forces is given by the Coulombs law through the use of the formula F=kq1q2/r2.0.1newtons = 8.99*109*3.2*10-6*7.7*10-7/r2 R= 555.78 Answer to question 3 • Potential difference between the two plates is equal to velocity which is equal to 6.0*106m/s • Force = mass *acceleration = 1.4*10-13*6.0*106 = -8254 nektons The speed of the particles are computed by the formula V=ED. This is equal to 8.5*10-6*0.15. This is equal to 84.1 Answer to question 4 Voltage = current *resistance. This implies that in this case while V is 5.0 and resistance is 1.0*103, current will be equal to 5/1.0*103, = 500 amps B the direction of the conventional current provides the electric charge movement from the positive side of the battery to its negative side as in indicated in the diagram below Answer to question 5 • This section focuses on the equivalent resistance of a circuit. The equivalent resistance will be equal to (5.0*102+1.00*103)2.
Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium. The reaction is as follows: Mg + 2HCl -> H2 + Mg2+ (aq) + 2Cl- (aq) There is a one to one relationship between the number of moles of hydrogen gas evolved and the
4Below is the high resolution 1H NMR spectrum of Compound G [pic] This compound contains C, H and N atoms. 1 How many peaks are there in the first signal (past the TMS from the right) and what can be concluded from this? 1½ marks 2 How many peaks has the second signal split into, and what can be concluded from this? 1½ marks 3 Determine the semi-structural formula of compound G and name the compound. 2 marks 4 Draw the structural formula of Compound G. 1 mark 5 Using the chemical shift correlation for 13C NMR, predict the number of peaks for Compound G and draw in the position of the peaks on the blank spectrum below, annotating each peak with its corresponding structure.
* Tape measure, 3 m * Mercury thermometer * Tuning fork, 384 Hz * Marker pencil * Block of wood C. Data Table 1: Tuning fork frequency (Hz) | Length, L Water level to top of the tube (m) | D= diameter of tube (m) | Wavelength=4(L+0.3d)(m) | Room temperature (degrees C) | 384 | 0.218 | 0.020 | 0.896 | 24 | D. Calculations A. In order to find the wavelength: Wavelength=4(L+0.3d) Next, plug in the values from Data Table 1 in to find the sound wavelength: Wavelength=4(0.218m +0.3*0.020m) Wavelength=4(0.218m+0.006m) Wavelength=4(0.224m) Wavelength=0.896 m In order to find the experimental value of v (speed of sound): V=f*wavelength (The value of f [frequency of the tuning fork] was found written on the side of the tuning fork.) v=384 Hz*.896m v=344.064 m/s In order to find actual speed of sound: (where Tc=room temperature in degrees Celsius) V speed of sound=331.4+0.6Tc m/s V speed of sound=331.4+0.6(24) m/s V speed of sound=345.8 m/s B. % error was found by using the following calculations: % error=(experimental value-theoretical value*100)/theoretical value % error= (344.064 m/s-345.8 m/s*100)/345.8 m/s % error=+-0.502% C. Consider the length of your resonance tube, what is the lowest frequency tuning fork you could use for this exercise? Show your calculations.
What is the oxidation number of S in SO2? A) +2 B) +4 C) 0 D) -2 E) +6 26. When the equation is properly balanced, the coefficient in front of H2O is: A) 22 B) 6 C) 0 D) 11 E) 4 27. When the equation is properly balanced, the coefficient in front of H+ is: A) 22 B) 6 C) 0 D) 11 E) 10 -------------------------------------------Please continue ------------------------------------- Balance the following REDOX reaction that is carried out under basic conditions. Show all work and label the two half-reactions by placing LEO and GER above the appropriate arrows: Answer the following questions that pertain to the reaction above and mark your answers on the Scantron ticket (questions 28 – 37): 28.
0.10 M NaCN can only combine with 1/2 as many Ag+'s when it makes (Ag(CN)2-) so, yoiur answer is really 0.05 molar solubility of Agl in 0.10 M NaCN ======================================… 3) Predict whether solubility of BaF2 will increase, decrease, or remain the same with the addition of: a) HCl adds H+ , which removes F- ion, by converting it into HF the equilib: BaF2 <=> Ba+2 & 2F- will shift to the right to replace lost F- so,solubility of BaF2 will increase b) KF adds F-, which increases the F- ion, the equilib: BaF2 <=> Ba+2 & 2F- will shift to the left to reduce the added F- so,solubility of BaF2 will decrease c) NaNO3 solubility of BaF2 will remain the same, Na+ions & NO3-1 ions are not in the equilibrium ... and they do not combine with, to remove either Ba+2 ions or F- ions d) Ba(NO3)2 adds Ba+2, which increases the Ba+2 ion concentration, the equilib: BaF2 <=> Ba+2 & 2F- will shift to the left to reduce the added
A second point charge q2 = -4.30 µC moves from the point x = 0.150 m, y = 0 to the point x = 0.250 m, y = 0.250 m. How much work is done by the electric force on q2? W (U b U a ) Ua Ub kq1q2 9(109 )(2.4)(106 )( 4.3)(106 ) 0.6192 J x1 0.150 kq1q2 x2 2 y2 2 9(109 )(2.4)(106 )( 4.3)(106 ) 0.252 0.252 0.2627 J W (U b U a ) ((0.2627) (0.6192)) 0.3565 J 2) Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge? KEi U i KE f U f KE f U i U
Concept Explorations 8.29. Periodic Properties I A hypothetical element, X, has the following ionization energy values: First ionization energy: 900 kJ/mol Second ionization energy: 1750 kJ/mol Third ionization energy: 14,900 kJ/mol Fourth ionization energy: 21,000 kJ/mol Another element, Y, has the following ionization energy values: First ionization energy: 1200 kJ/mol Second ionization energy: 2500 kJ/mol Third ionization energy: 19,900 kJ/mol Fourth ionization energy: 26,000 kJ/mol * a. To what family of the periodic table would element X be most likely to belong? Explain? * X belongs to Group 2 of the periodic table because it has a low first and second ionization energy, however, a higher second and third.