1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Aneli [31]
3 years ago
15

Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.

54 cm. (1 f-12 inches)
Chemistry
1 answer:
Jlenok [28]3 years ago
6 0

<u>Answer:</u> The density of gold in g/cm^3 is 19.22g/cm^3

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}

We are given:

Density of gold = 1200lb/ft^3

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into g/cm^3, we get:

\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3

Hence, the density of gold in g/cm^3 is 19.22g/cm^3

You might be interested in
For each pair of bonds, indicate the more polar bond, and use an arrow to show the direction of polarity in each bond.a. C-O and
Naddika [18.5K]

Answer:

The more polar bond is C-O.

Explanation:

The greater polarity is due to that the Oxygen atom is more electronegative than the Nitrogen atom, so the negativity of the molecule tends to be on that side.

7 0
1 year ago
If cats take over the world, will you make them extinct or be their slave.
prisoha [69]

Answer:

Make them go instinct

Explanation:

No one wants cats taking over the whole world

3 0
3 years ago
Which models are most appropriate for him to use?
jekas [21]

Answer:

the triangular model and the cubic model

5 0
3 years ago
Read 2 more answers
What is Avogadro's number, and why is it useful? (3 points)
SashulF [63]

Answer:

6.022 x 10²³; it is a conversion factor between moles and number of particles

Explanation:

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole of hydrogen  = 6.022 × 10²³ atoms of hydrogen

238 g of uranium = 1 mole of uranium = 6.022 × 10²³ atoms of uranium

By taking ions:

62 g of NO⁻₃ =  1 mole of  NO⁻₃  = 6.022 × 10²³ ions of  NO⁻₃

96 g of SO₄²⁻ = 1 mole of SO₄²⁻ =  6.022 × 10²³ ions of  SO₄²⁻

4 0
3 years ago
The chemical formula for sulfuric acid is H2SO4. In one molecule of sulfuric acid, there are _____.
erma4kov [3.2K]

Answer:

four atoms of oxygen

Explanation:

two atoms of hydrogen

one sulphur atom

5 0
3 years ago
Read 2 more answers
Other questions:
  • A reaction expected to produce 245.6 g of a product produces 122.1 g of the
    7·1 answer
  • A urine sample has a mass of 147 g and a density of 1.010 g/ml. what is the volume of this sample?
    8·1 answer
  • What qualities make one theory "better" than another in the scientific community?
    6·1 answer
  • Young's experiment is performed with light from excited helium atoms (λ = 496 nm). fringes are measured carefully on a screen 1.
    6·1 answer
  • A tank of water contains many billions of water molecules. These molecules move at random speeds and in random directions, often
    10·1 answer
  • Use your periodic table to answer the following questions. Pay attention to the spelling of the element names and make sure to w
    15·1 answer
  • 1. Hg(OH)2 + Ag →<br> what is the answer for this?
    6·1 answer
  • I will give 15 points
    15·2 answers
  • A scientist observes that the electrical resistance of a superconducting material drops to zero when the material is cooled
    9·1 answer
  • Need help on this question
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!