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Alchen [17]
3 years ago
10

An empty vial weighs 55.32 g. (a) If the vial weighs 185.56 g when filled with liquid mercury (d = 13.53 g/cm3). What i its volu

me? (b) How much would the vial weigh if it were filled with water (d= 0.997 g/cm3 at 25°C)?
Chemistry
1 answer:
Tasya [4]3 years ago
3 0

Answer:

a) Volume of vial= 9.626cm3

b) Mass of vial with water = 62.92 g

Explanation:

a) Mass of empty vial = 55.32 g

Mass of Vial + Hg = 185.56 g

Therefore,

mass\ of\ Hg = 185.56-55.32 = 130.24 g

Density of Hg = 13.53 g/cm3

Volume\ of\ vial = Volume\ of\ Hg = \frac{Mass}{Density} \\\\= \frac{130.24g}{13.53g/cm3} = 9.626 cm3

b) Volume of water = volume of vial = 9.626 cm3

Density of water = 0.997 g/cm3

Mass\ of\ water = Density*volume = 0.997g/cm3*9.626cm3=9.60 g\\\\Total\ Mass\ of\ vial = Empty\ vial + mass\ of\ water\\= 53.32+9.60= 62.92g

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How many atoms of S are there?
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The piece of iron that miguel measured had a mass of 51.1 g and a volume of 6.63 cm 3 . what did miguel calculate to be the dens
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Given:
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In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
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A 8.65-L container holds a mixture of two gases at 11 °C. The partial pressures of gas A and gas B, respectively, are 0.205 atm
Vsevolod [243]

 The  total pressure  = 1.402 atm


<u><em>calculation</em></u>

Total  pressure = partial  pressure  of gas A + partial pressure of gas B +  partial pressure  of third gas

partial  pressure  of gas A= 0.205 atm

Partial pressure of gas B =0.658 atm


partial pressure for third gas is calculated using ideal  gas equation

that is PV=nRT   where,

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=1.402 atm

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3 years ago
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