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Minchanka [31]
3 years ago
8

An ice cube of volume 15 cm^3 and density 917 kg/m^3 is placed in ethyl alcohol of density 811 kg/m^3. What is the buoyant force

on the ice? Will it float?
Physics
1 answer:
Amiraneli [1.4K]3 years ago
3 0

Answer:

the buoyant force acting on the  ice cube is 0.119 N

Explanation:

given,

volume of ice cube = 15 cm³ = 15 × 10⁻⁶ m³

density of ice cube = 917 kg/m³

density of ethyl alcohol = 811 kg/m³

buoyant force = ?

The density of ice is more than ethyl alcohol hence it will sink.

buoyant  force acting on the ice cube = ρ V g

                                                                = 811 × 15 × 10⁻⁶ × 9.81

                                                                =0.119 N

so, the buoyant force acting on the  ice cube is 0.119 N

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onsider 1000 mL of a 1.00 × 10-4 M solution of a certain acid HA that has a Ka value equal to 1.00 × 10-4. Water was added or re
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Number of moles of HA =  1.00 * 10⁻⁴ mols

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HA → H⁺  +  A⁻

If 1 mol of HA produces 1 mol of H⁺  and  A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of  H⁺  and  A⁻.

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Number of moles of  H⁺ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  H⁺ produced = 1.6 * 10⁻⁵mols

Number of moles of  A⁻ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  A⁻ produced = 1.6 * 10⁻⁵mols

Since 16% of HA dissociated into  H⁺  and  A⁻, 84% of HA is left

Number of mols of HA left = 0.84 *  1.00 * 10⁻⁴

Number of mols of HA left =  8.4 * 10⁻⁵mols

Concentration = num of moles/volume

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Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

Conc of A⁻ =  1.6 * 10⁻⁵/V

To calculate the dissociation constant

k_{a} = [H^{+} ][A^{-} ]/[HA]

k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L

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