Answer:
99.24%.
Explanation:
- NaCl reacted with AgNO₃ as in the balanced equation:
<em>NaCl + AgNO₃ → AgCl(↓) + NaNO₃,</em>
1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.
- We need to calculate the no. of moles of AgCl produced:
no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.
- Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:
<em>using cross multiplication:</em>
1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.
∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.
- Now, we can get the mass of puree NaCl in the sample:
mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.
∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.
Because the alkali metals are the group 1 metals, they have only 1 valence electron that they want to lose, and the halogens are the group 17 nonmetals, they want to gain 1 valence electron to become stable.
Answer:
0.077M is the concentration of the hydroxyl ion
Explanation:
Dilution factor is the ratio between the aliquot that is taken of a solution and the total volume of the diluted solution.
For the problem, dilution factor is:
7.53cm³ / 147cm³ =<em> 0.05122</em>
To obtain molarity of a diluted solution you must multiply dilution factor and initial molarity of the solution, thus:
1.5 M × 0.05122 = <em>0.077M is the concentration of the hydroxyl ion</em>
Answer:
1s2 2s2 2p2
Explanation:
it has 6 electrons in two energy levels so the sub levels are 1s, 2s and 2p
