The relative mass of each element can be found from the periodic table (the larger number). For instance, P2O5, P = 31.0 O = 16.0, thus the formula mass (Mr) is 2(31) + 5(16) = 142 amu (atomic mass unit). I used a not really specific periodic table. Maybe your teacher is referring to open Schoology (a website where teacher can post files or announcements to students in his or her class). Sorry if i got something mistaken.
Answer:
(c) The retention time would be higher (d) The retention time would be lower.
Explanation:
For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.
However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.
Answer:
1) 1.235 g.
2) 0.61 g.
Explanation:
- From the balanced equation:
<em>Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O.</em>
1.0 mol of Al(OH)₃ reacts with 3.0 moles of HCl to produce 1.0 mol of AlCl₃ and 3.0 moles of H₂O.
<em>1) How many grams of HCl can a tablet with 0.880 g of Al(OH)₃ consume? </em>
- To calculate the amount of HCl needed to consume 0.880 g of Al(OH)₃, we need to calculate the no. of moles of Al(OH)₃:
no. of moles of Al(OH)₃ = mass/molar mass = (0.880 g)/(78.0 g/mol) = 1.13 x 10⁻² mol.
∵ Every 1.0 mol of Al(OH)₃ needs 3.0 moles of HCl to be consumed.
∴ 1.13 x 10⁻² mol of Al(OH)₃ needs (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of HCl.
The no. of grams of HCl = no. of moles of HCl x molar mass of HCl = (3.385 x 10⁻² mol)(36.5 g/mol) = 1.235 g.
<em>2) How much H₂O?</em>
∵ Every 1.0 mol of Al(OH)₃ produces 3.0 moles of H₂O.
∴ 1.13 x 10⁻² mol of Al(OH)₃ produces (3 x 1.13 x 10⁻² = 3.385 x 10⁻² mol) of H₂O.
<em>The no. of grams of H₂O = no. of moles of H₂O x molar mass of H₂O </em>= (3.385 x 10⁻² mol)(18.0 g/mol) = <em>0.6092 g ≅ 0.61 g.</em>
