The sulphate solutions came from a recycling LIBs waste cathode materials, which were done by previous research; their content is shown in Table 1 [18]. Sodium carbonate (Na2CO3) was purchased from Nihon Shiyaku Reagent, Tokyo, Japan (NaCO3, 99.8%), for the chemical precipitation. CO2 was purchased from Air Product and Chemical, Taipei, Taiwan (CO2 ≥ 99%), to carry out the hydrogenation–decomposition method. Dowex G26 was obtained from Sigma-Aldrich (St. Louis, MO, USA) and was used as a strong acidic cation exchange resin, to remove impurities. Multi-elements ICP standard solutions were acquired from AccuStandard, New Haven, Connecticut State, USA. The nitric acid (HNO3) and sulfuric acid (H2SO4) were acquired from Sigma-Aldrich (St. Louis, MO, USA) (HNO3 ≥ 65%) (H2SO4 ≥ 98%) The materials were analyzed by energy-dispersive X-ray spectroscopy (EDS; XFlash6110, Bruker, Billerica, MA, USA), X-ray diffraction (XRD; DX-2700, Dangdong City, Liaoning, China), scanning electron microscopy (SEM; S-3000N, Hitachi, Tokyo, Japan), and inductively coupled plasma optical emission spectrometry (ICP-OES; Varian, Vista-MPX, PerkinElmer, Waltham, MA, USA). In order to
Appl. Sci. 2018, 8, 2252 3 of 10
control the hydrogenation temperature and heating rate, a thermostatic bath (XMtd-204;
Answer:
Silver ions 1 Hydrogen ions 2 Iron ions 3 Sodium ions 4
Explanation:
Answer:
THE FREEZING POINT OF A WATER SOLUTION OF FRUCTOSE MADE BY DISSOLVING 92 g OF FRUCTOSE IN 202g OF WATER IS -4.70 ◦C
Explanation:
To calculate the freezing point of a water solution of fructose,
1. calculate the molar mass of Fructose
( 12 * 6 + 1*12 + 16*6) =72 + 12 +96 = 180 g/mol
2. calculate the number of moles of fructose in the solution
number of moles = mass / molar mass
n = 92 g / 180 g/mol
n = 0.511 moles.
3. calculate the molarity of the solution
molarity = moles / mass of water in kg
molarity = 0.5111 / 202 g /1000 g
molarity = 0.5111 / 0.202
molarity = 2.529 M
4. calculate the change in the freezing point of pure solvent and solution ΔTf
ΔTf = Kf * molarity of the solute
Kf = 1.86 ◦C/m for water
ΔTf = 1.86 * 2.529
ΔTf = 4.70 C
5. the freezing point is therefore
0.00 ◦C - 4.70 ◦C = -4.70 ◦C
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