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svetlana [45]
2 years ago
8

The density (mass/volume) of aluminum is 2.70 times. 103 kilograms per cubic meter (kg/m3). what is the mass of an aluminum cyli

nder that has a volume of 1.50 m3? 5.56 times. 10–4 kg 1.50 times. 10–3 kg 1.80 times. 103 kg 4.05 times. 103 kg
Chemistry
1 answer:
wariber [46]2 years ago
7 0

Mass of aluminum = (1.50 m^3)( 2.70 x 10^3 kg/m^3) = 4050 kg 

The intention of density is the assembly per volume of the substance. This is an intrinsic property consequently the size (or any external feature) does not change this effects.

<h3>What is Aluminum?</h3>
  • Aluminum is a silvery-white metal, the 13 elements in the regular table. One surprising fact regarding aluminum is that it's the most widespread metal on Earth, making up more than 8% of the Earth's core mass. It's also the third most standard chemical essence on our planet after oxygen and silicon.
  • The formed aluminum is in everyday use in mining, manufacturing, and trade in the United States; the formed aluminum is used with reasonable uniformity in Great Britain and typically by chemists in the United States.
  • Aluminum (Al), also spelled aluminum, chemical component, a weightless silvery-white metal of main Group 13 (IIIa, or boron group) of the periodic table. Aluminum is the most plentiful metallic component in Earth's crust and the numerous widely used nonferrous metal.

To learn motrev about  Aluminum, refer to:

brainly.com/question/246454

#SPJ4

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<u>Answer:</u> The freezing point of solution is -0.454°C

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Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

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m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

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