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r-ruslan [8.4K]
3 years ago
6

You measure salt water in a tank to have a density of 1.02 g/mL. A balloon weighs 2.0 g and you weights have a mass of 30.0 g ea

ch. If you put 6 weights in your balloon, you must inflate your balloon to what diameter for it to have a density equal to the salt water, and therefor float in the middle of the tank?
Chemistry
1 answer:
Elden [556K]3 years ago
4 0

Weight of the balloon = 2.0 g

Six weights each of mass 30.0 g is added to the balloon.

Total mass of the balloon = 2.0 g + 6*30.0 g = 182 g

Density of salt water = 1.02 g/mL

Calculating the volume from mass and density:

182g*\frac{mL}{1.02g} =178mL

Converting the volume from mL to cubic cm:

178 mL * \frac{1cm^{3} }{1mL} =178cm^{3}

Assuming the balloon to be a sphere,

Volume of the sphere = \frac{4}{3}πr^{3}

178 cm^{3} = \frac{4}{3}(\frac{22}{7})r^{3}

r = 3.49 cm

Radius of the balloon = 3.49 cm

Diameter of the balloon = 2 r = 2*3.49 cm = 6.98cm


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Answer:

The correct answer is: Ka= 5.0 x 10⁻⁶

Explanation:

The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:

               HA               ⇄        H⁺        +          A⁻

t= 0      0.200 M                     0                     0

t              -x                             x                       x

t= eq      0.200M -x               x                       x

At equilibrium, we have the following ionization constant expression (Ka):

Ka= \frac{ [H^{+}]  [A^{-} ]}{ [HA]}

Ka= \frac{x x}{0.200 M -x}

Ka= \frac{x^{2} }{0.200 M - x}

From the definition of pH, we know that:

pH= - log  [H⁺]

In this case, [H⁺]= x, so:

pH= -log x

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⇒x = 10⁻³

We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:

Ka= \frac{(10^{-3})^{2}  }{0.200 - (10^{-3}) }= \frac{10^{-6} }{0.199}= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶

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