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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A compound is two or more chemically combined elements. Fore example NaCl, CO2, H2O
The answer is C. because <span>particles settle out over time ,can block light and scatter light .</span>
Answer:
The coefficient before potassium (K) balances this chemical equation is 2.
Explanation:
_K +Cl₂ → 2KCl
K =1 ; Cl =2
K=1 × 2 = 2
Cl = 1 × 2 = 2
2 K +Cl₂ = 2 KCl
Answer : Option 1) The true statement is each carbon-oxygen bond is somewhere between a single and double bond and the actual structure of format is an average of the two resonance forms.
Explanation : The actual structure of formate is found to be a resonance hybrid of the two resonating forms. The actual structure for formate do not switches back and forth between two resonance forms.
The O atom in the formate molecule with one bond and three lone pairs, in the resonance form left with reference to the attached image, gets changed into O atom with two bonds and two lone pairs.
Again, the O atom with two bonds and two lone pairs on the resonance form left, changed into O atom with one bond and three lone pairs. It concludes that each carbon-oxygen bond is neither a single bond nor a double bond; each carbon-oxygen bond is somewhere between a single and double bond.
Also, it is seen that each oxygen atom does not have neither a double bond nor a single bond 50% of the time.