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Rudik [331]
3 years ago
9

How could you predict the sinking/floating behavior of a spherical object?

Chemistry
2 answers:
antiseptic1488 [7]3 years ago
8 0
You can predict it by measuring the weight of the spherical object and measuring the liquid (mL) and subtracting finding the volume
worty [1.4K]3 years ago
3 0

To predict the sinking/floating behavior of a spherical object, you could determine its density.

<em>Step 1</em>. Determine its <em>mass</em> on a suitable balance.

<em>Step 2</em>. Determine its <em>volume</em>

<em>Either </em>

Measure its circumference. <em>C</em> = 2π<em>r</em>, so you can calculate r

<em>or </em>

Measure its diameter. <em>d</em> = 2<em>r</em>, so you can calculate <em>r </em>

<em>Then </em>

<em>V</em> = ⁴/₃π<em>r</em>³ , so you can calculate its volume.

<em>Step 3</em>. Calculate its <em>density </em>(<em>D</em>).

Density = mass/volume

If D < 1 g/cm³, the object will float in water. If D > 1 g/cm³, the object will sink.

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How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?
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The number  of Ml  of  a  0.40 %w/v solution  of   ,nalorphine  that must  be injected  to  obtain  a  dose  of 1.5 mg is  calculated as  below


since M/v%   is  mass  of solute  in  grams per 100  ml

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6 0
3 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
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