Since a water molecule is H2O, you would divide 126 hydrogen molecules by 2, and you would get 63. That means you have 63 double hydrogen molecules, and 58 oxygen molecules to pair up with them. So that means you could have 58 molecules of water, with 5 double hydrogen molecules, so basically 10 extra molecules of hydrogen along with the H2O molecules. Hope I helped! :)
Cations from smallest to largest
Li⁺ ,Na⁺, K⁺ (from Periodic Table, the bigger number of period, the bigger size, of atom, so the bigger size of cation)
1) LiF smaller cation then KF
1,036 <span>853
</span><span>The lattice energy increases as cations get smaller, as shown by LiF and KF.
</span><span>I think this one should be correct answer, because the compared substances have also the same anion, and we can compare cations in them.
2) The same cation Li , so wrong statement.
3)</span>The same cation Na , so wrong statement.
4) NaCl smaller cation then KF
786 853
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%.