<span>6.12<span>(<span>1024</span>)</span></span><span>=<span><span>(6.12)</span><span>(<span><span>1e</span>+24</span>)</span></span></span><span>=<span><span>6.12e</span>+24</span></span>
<span>
=
</span>
Answer:
(119 g H2O) / (18.01532 g H2O/mol) x (1 mol Pb / 2 mol H2O) x (207.21 g Pb/mol) = 684 g Pb
Explanation:
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
Grams of Phosphorus = 4.14 grams
Grams of white compound = 27.8 grams
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
Calculating moles which would be grams / molar mass
Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
Calculating the ratios by dividing with the small entity
P = 0.1337 moles / 0.1337 moles = 1
Cl = 0.6674 moles / 0.1337 moles = 5
So the empirical formula would be PCl5
54.868% because well la la la (sorry had to be 20 characters)
