Answer:
A. 8.4
Explanation:
[OH⁻] = 2.6 × 10⁻⁶ Take the negative log of each side
-log[OH⁻] = pOH = 5.59 Apply the pH/pOH relation
pH + pOH = 14.00 Insert the value of pOH
pH + 5.59 = 14.00 Subtract 5.59 from each side
pH = 14.00 -5.59 = 8.41
Answer is: detergent is basic.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.
When pH is less than seven (pH<7), solution is acidic.
When is equal seven (pH = 7), solution is neutral.
When pH is greater than seven (pH > 7), like in this example, solution is basic.
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>
Answer:
32.8%
Explanation:
All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).
First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:
- 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄
There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.
We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:
- 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb
Finally we calculate the percentage composition of Pb:
- 0.083 g Pb / 0.254 g salt * 100% = 32.8%
Answer:
That should be Hydrogen.
So The second option Is legit!
