Answer:
In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams
Explanation:
This compound is the lead (IV) phosphate.
Grams that occupy 1 mole, means the molar mass of the compound
Pb = 207.2 .3 = 621.6 g/m
P = 30.97 .4 = 123.88 g/m
O = (16 . 4) . 4 = 256 g/m
621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m
Answer:
Then, at some point, these higher energy electrons give up their "extra" energy in the form of a photon of light, and fall back down to their original energy level.
Explanation:
When properly stimulated, electrons in these materials move from a lower level of energy up to a higher level of energy and occupy a different orbital.
The complete balanced chemical equation for this is:
<span>3KOH + H3PO4
--> K3PO4 + 3H2O</span>
First we calculate the number of moles of H3PO4:
moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol
From stoichiometry, 3 moles of KOH is required for every
mole of H3PO4, therefore:
moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole
H3PO4) = 0.0468 mol
Calculating for volume given molarity of 0.350 M KOH:
Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7
mL
Answer:
<span>133.7 mL KOH</span>
Answer:
Explanation:
The molecular mass of C2H6 is approximately 30 or [(2 x 12) + (6 x 1)]. Therefore the molecule is about 2.5 times as heavy as the 12C atom or about the same mass as the NO atom with a molecular mass of 30 or (14+16).
