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Furkat [3]
1 year ago
6

You plan to use the water displacement method to determine if a ring is pure silver

Chemistry
1 answer:
timama [110]1 year ago
6 0

The water displacement method is a set of measurements used to calculate the volume of an irregularly shaped object. These objects are commonly known as irregularly shaped solids.

<h3>What is water displacement method ?</h3>

The displacement method (also known as the submersion or dunking method) can be used to accurately measure the volume of the human body and other irregularly shaped objects by measuring the volume of fluid displaced when the object is submerged.

Place your jewelry on a table or in your hand, and pour some white vinegar directly on the metal (a dropper can also be used). If the metal of the jewelry changes color, it is not pure gold; if it continues to shine, you have real gold in your hand.

Thus, The water displacement method is a set of measurements used to calculate the volume of an irregularly shaped object.

To learn more about water displacement method, follow the link;

brainly.com/question/17342316

#SPJ1

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A 29.00 mL sample of an unknown H3PO4 solution is titrated with a 0.130 M NaOH solution. The equivalence point is reached when 2
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A 18.08-g sample of the ionic compound , where is the anion of a weak acid, was dissolved in enough water to make 116.0 mL of so
Oksi-84 [34.3K]

Answer:

a) 129.14 g/mol

b) 8.87

Explanation:

Given that:

mass of the ionic compound [NaA] = 18.08 g

Volume of water = 116.0 mL = 0.116 L

Let the mole of the acid HCl = 0.140 M

Volume of the acid = 500.0 mL = 0.500 L

pH = 4.63

V_{equivalence}_{acid} = 1.00 L

Equation for the reaction can be represented as:

NaA_{(aq)} + HCl_{(aq)} -----> HA_{(aq)} + NaCl_{(aq)

From above; 1 mole of an ionic compound reacts with 1 mole of an acid to reach equivalence point = 0.140 M × 1.00 L

= 0.140 mol

Thus, 0.140 mol of HCl neutralize 0.140 mol of ionic compound at equilibrium

Thus, the molar mass of the sample = \frac{18.08g}{0.140 mole}

= 129.14 g/mol

b) since pH = pKa

Then pKa of HA = 4.63

Ka = 10^{-4.63]

= 2.3*10^{-5}

[A^-]equ = \frac{0.140M*1.00L}{1.00L+0.116L}

= \frac{0.140 mol}{1.116 L}

= 0.1255 M

K_a of HA = 2.3*10^{-5}

K_b = \frac{1.0*10^{-14}}{2.3*10^{-5}}

= 4.35*10^{-10}

                     A_{(aq)}     +     H_2O_{(l)}         \rightleftharpoons     HA_{(aq)}     +     OH^-_{(aq)}

Initial        0.1255                                            0                    0

Change     - x                                                  +  x                 + x

Equilibrium   0.1255 - x                                   x                    x

K_b = \frac{[HA][OH^-]}{[A^-]}

4.35*10^{-10} = \frac{[x][x]}{[0.1255-x]}

As K_b is very small, (o.1255 - x) = 0.1255

x = \sqrt{0.1255*4.35*10^{-10}}

[OH⁻] = x = 7.4 *10^{-6}

But pOH = - log [OH⁻]

= - log [7.4*10^{-6}]

= 5.13

pH = 14.00 = 5.13

pH = 8.87

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3 years ago
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