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Vikentia [17]
3 years ago
14

Suppose you are titrating an acid solution with a base solution of known concentration. To calculate the concentration of the ac

id solution, use three steps.
1. Use the -------------- { (a) delivered volume (b) volume reading} of base --------------- { (a) after (b) to reach (c) before} the endpoint and the known concentration of the base solution to find the ------------ { (a) moles (b) mass (c) concentration} of base used.
2. Use the ------------ { (a) molar mass (b) dilution equation (c) molar ratio} to find the moles of acid from the moles of base.
3. Divide the moles of acid by the volume of -------------- { (a) base solution before (b) acid solution before (c) acid solution after} the titration to find the concentration of acid.
Chemistry
1 answer:
Arturiano [62]3 years ago
4 0

Answer:

i would say use 2

Explanation:

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Sodium hydroxide, NaOHNaOH; sodium phosphate, Na3PO4Na3PO4; and sodium nitrate, NaNO3NaNO3, are all common chemicals used in cle
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Answer:

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Explanation:

in NaOH

23/40 * 100 = 57.5%

in Na3PO4

3 * 23/164 * 100 = 42%

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23/85 * 100 = 27.1%

Hence;

NaOH > Na3PO4 > NaNO3

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8 0
3 years ago
Read 2 more answers
A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated cylin
gtnhenbr [62]

Answer:

ksp = 0,176

Explanation:

The borax (Na₂borate) in water is in equilibrium, thus:

Na₂borate(s) ⇄ borate²⁻(aq) + 2Na⁺(aq)

<em>When you add just borax, the moles of Na²⁺ are twice the moles of borate²⁻, that means 2borate²⁻=Na⁺ </em><em>(1)</em>

The ksp is defined as:

<em>ksp = [borate²⁻] [Na⁺]²</em>

Then, borate²⁻(B₄O₇²⁻) reacts with HCl thus:

B₄O₇²⁻ + 2HCl + 5H₂O → 4H₃BO₃ + 2Cl⁻

The moles of HCl that reacts with B₄O₇²⁻ are:

0,500M×0,01200L = 6,00x10⁻³ mol of HCl

As two moles of HCl react with 1 mol of B₄O₇²⁻, the moles of B₄O₇²⁻ are:

6,00x10⁻³ mol of HCl×\frac{1molB_{4}O_{7}^{2-}}{2molHCl} = <em>3,00x10⁻³ mol of B₄O₇²⁻</em>

For (1), moles of Na⁺ are <em>3,00x10⁻³ mol ×2 = 6,00x10⁻³ mol of Na⁺</em>

The [borate²⁻] is <em>3,00x10⁻³ mol of B₄O₇²⁻/0,00850L = </em><em>0,353M</em>

And [Na⁺] is <em>6,00x10⁻³ mol of Na⁺ / 0,00850L = </em>0,706M

Replacing in the expression of ksp:

ksp = [0,353] [0,706]²

<em>ksp = 0,176</em>

<em></em>

I hope it helps!

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To find out if the height from which a student drops a ball affects how high the ball bounces, the independent variable is the
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The independent variable is the height of the ball that bounces.

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