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loris [4]
3 years ago
15

A block of metal has a mass of 18.361kg and the following dimensions: 22.35 cm x 7.34 cm x 22.05 mm. What is the density of the

metal in g/cm^3
Chemistry
1 answer:
professor190 [17]3 years ago
3 0

Answer:

50.76 g/cm³

Explanation:

We'll begin by converting 18.361 Kg to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

18.361 Kg = 18.361 Kg × 1000 g / 1 Kg

18.361 Kg = 18361 g

Next, we shall express the dimension in the same unit of measurement.

Dimension = 22.35 cm x 7.34 cm x 22.05 mm

Thus, we shall convert 22.05 mm to cm. This can be obtained as follow:

10 mm = 1 cm

Therefore,

22.05 mm = 22.05 mm × 1 cm / 10 mm

22.05 mm = 2.205 cm

Thus, the dimension is 22.35 cm x 7.34 cm x 2.205 cm.

Next, we shall determine the volume of the metal. This can be obtained as follow:

Dimension = 22.35 cm x 7.34 cm x 2.205 cm.

Volume = 361.728045 cm³

Finally, we shall determine the density of the metal. This can be obtained as follow:

Mass of metal = 18361 g

Volume of metal = 361.728045 cm³

Density of metal =?

Density = mass / volume

Density of metal = 18361 / 361.728045

Density of metal = 50.76 g/cm³

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Explanation:

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Why do substances with different structures have similar conductivity values
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Answer:

because of the material, they are made of      

Explanation:

like a paperclip and a car door maybe they both are metal or steel whatever they both can conduct energy. because of what they are made of.

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7 0
3 years ago
What is the volume of 40.0 grams of argon gas at STP ?
MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Aqueous Solutions</u>

  • States of Matter

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 40.0 g Ar

[Solve] L Ar

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 24.9235 \ L \ Ar

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

24.9235 L Ar ≈ 24.9 L Ar

5 0
3 years ago
What mass of H2SO4 is contained in 60.00 mL of a 5.85 M solution of sulfuric acid
strojnjashka [21]

First, we have to calculate the number of moles of H2SO4 in the solution:

V=60 mL = 0.06 L

c=5.85 mol/L

n=V×c=0.06×5.85=0.351 mol

Then we need to find the molar mass of H2SO4:

2×Ar(H) + Ar(S) + 4×Ar(O) =

=2 + 32 + 64 = 98 g/mol

Finally, we need to find the mass of H2SO4:

m=0.351 × 98 = 34.398 g

5 0
3 years ago
What identifies a flaw in J. J. Thomson’s model of the atom?
katrin [286]

Answer:

The plum-pudding model did not identify a central nucleus as the source of a positive charge.

Explanation:

Sir Joseph John Thomson, most popularly known as J J Thomson. He was a famous scientist who was awarded the Noble prize for Physics for his discovery of the subatomic particle, electron.

He placed his famous model of an atom which is known as plum pudding model. He proposed that the atoms are described as the negative particles that is floating within a soup of the diffuse positive charge.

But the main defect of his proposal was that it did not recognized the presence of a central nucleus as a positive charged source.

3 0
3 years ago
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