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2 years ago

What is a change that will not affect the pressure in a container?

Chemistry

1 answer:

allochka39001 [22]2 years ago

7 0

Answer:

Density of the contents

Explanation:

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2 years ago

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

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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 = $\frac{4}{3}$ π $r^{3}$

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

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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3 years ago

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2 years ago

What is a change that will not affect the pressure in a container?​

What is a change that will not affect the pressure in a container?