Cl, because a higher electronegativity indicates that the atom attracts more electrons.
Moles of P = 56,4g/30,974g/mole = 1,82 moles P
moles of O = 43,6/15,999 = 2,73 moles of O
converting to the simplest ratio:
For P : 1,82/1,82 = 1
For O : 2,73/1,82 = 1,5
1 P and 2 oxygens.
PO2 -> the empirical formula
hope this help
We need to first calculate the empirical formula. Empirical formula is the simplest ratio of whole numbers of components in a compound,
Mass percentages have been given. We need to then calculate for 100 g of the compound
C H O
mass 77.87 g 11.76 g <span>10.37 g
number of moles 77.87/12 11.76/1 10.37/16
moles = 6.48 = 11.76 =0.648
divide by least number of moles
6.48/0.648 11.76/0.648 0.648/0.648
= 10 =18.1 = 1
rounded off
C - 10 , H - 18 and O - 1
empirical formula - C</span>₁₀H₁₈O
mass of empirical unit = 12 x 10 + 1x 18 + 16 = 120 + 18 + 16 = 154
number of empirical units = molecular mass / mass of one empirical unit
= 154.25 / 154 = 1.00
Therefore molecular formula = C₁₀H₁₈O
Answer:
pH = 12.8
Explanation:
HF + NaOH → F⁻ + Na⁺ + H₂O
<em>1 mole of HF reacts with 1 mole of NaOH</em>
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Initial moles of HF and NaOH are:
HF = 0.018L × (0.308mol / L) = 5.544x10⁻³mol HF
NaOH = 0.023L × (0.361mol / L) = 8.303x10⁻³mol NaOH
That means moles of NaOH remains after reaction are:
8.303x10⁻³mol - 5.544x10⁻³mol = <em>2.759x10⁻³moles NaOH</em>
Total volume is 18.0mL + 23.0mL = 41.0mL = 0.0410L
Molar concentration of NaOH is
2.759x10⁻³moles NaOH / 0.0410L = 0.0673M = [OH⁻]
pOH = - log [OH⁻] = 1.17
As pH = 14 - pOH
<em>pH = 12.8</em>
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