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Vadim26 [7]
1 year ago
5

At room temperature (208C) and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

n that of air. In a buoyancy experiment with a new plastic, a chemist creates a rigid, thin-walled ball that weighs 0.12 g and has a volume of 560 cm⁹.
(e) Will it float if filled with nitrogen (d = 1.165 g/L)?
Chemistry
1 answer:
Ugo [173]1 year ago
4 0

Total density of filled ball with nitrogen gas: \frac{0.12G+0.6524g}{0.560L} = 1.3792g/L

The relationship between mass and volume can be easily determined using density; for example, the mass of a body is equal to its volume multiplied by the density (M = Vd), whereas the volume is equal to the mass divided by the density (V = M/d). The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air. Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = d_4=1.165g/L

Mass of the nitrogen gas : 1.165g/L \times 0.560L = 0.6524g

Learn more about Mass and Density here:

brainly.com/question/10821730

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This question in the screenshot
Morgarella [4.7K]

Mass of hydrate + crucible = 47.29 g

Mass of anhydrous salt = 2.7 g

Molar mass of anhydrous salt CuSO4 = 159.5 g

Given,

mass of empty crucible = 42.45 g

mass of hydrate salt= 4.84 g

mass of crucible after first heating = 46.1 g

mass of crucible after second heating= 45.153 g

mass of crucible after third heating= 45.15 g

so, as per the question we need to find...

Mass of hydrate + crucible = ? g

Mass of anhydrous salt = ? g

Molar mass of anhydrous salt CuSO4 = ? g

∴Mass of hydrate + crucible = 42.45 + 4.48 = 47.29 g

The given salt is in hydrate form, to remove water from this molecule we need to perform heating .

So we are taking the substance into the crucible as it is in less quantity.

Here, we performed heating 3 times and note the weight after every heating.

After this, assume that the water is totally evaporated and the remaining salt is in anhydrous form,

∴ Mass of anhydrous salt = 45.15 - 42.45 = 2.7 g

To find the molar mass of anhydrous salt of CuSO4,

atomic weight of Cu = 63.5 g

atomic weight of S = 32 g

atomic weight of O =16 g

∴ molar mass of anhydrous salt of CuSO4 = 63.5 + 32 + (16 ×3)

                                                                   =159.5 gm

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2 years ago
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