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cupoosta [38]
2 years ago
14

a 100 gram sample of iron ore was found to contain 44g of iron. How many grams of iron are in a 450 gram sample of the ore?

Chemistry
1 answer:
Sphinxa [80]2 years ago
7 0

Answer:

There are 198 grams of iron in a 450-gram sample of the ore.

Explanation:

From the given information:

Given that:

100-gram sample of iron ore is found in 44g of iron.

Thus. the grams of iron that can be found in a 450-gram sample of the iron ore can be computed as follows:

100g of iron ore = 44g of iron

450 gram of iron ore = x g of iron

x(g) of iron = ( 44g of iron × 450 gram of iron ore) / 100g of iron ore

x(g) of iron =  198 grams of iron

Thus, there are 198 grams of iron in a 450-gram sample of the ore.

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to become a noble gas element P will have 2 electrons in it's outer most energy level if it has one energy level

and eight in the last energy level if more than one

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A) How many grams of water are produced when 2.50 g of aluminum
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an atmosphere is considered hazardous if it contains a hazardous gas in excess of 10 percent of the hazardous material's:
svlad2 [7]

Lower flammable limit means the lowest concentration of a material that will propagate a flame.

What is hazardous atmosphere?

It is an atmosphere that may expose employees to risk of death, incapacitation, impairment of ability to self-rescue, injury, or acute illness from one or more of following causes

  • Flammable gas, vapor, or mist in excess of 10 percent of lower flammable limit (LFL)
  • Airborne combustible dust at concentration that meets or exceeds its LFL

What is lower flammable limit?

  • It means the lowest concentration of a material that will propagate a flame.
  • The LFL is usually expressed as percent by volume of material in air (or other oxidant)
  • Atmospheres with concentration of flammable vapors at or above 10 percent of lower explosive limit (LEL) are considered hazardous when located in confined spaces.
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Learn more about lower flammable limit at brainly.com/question/2456135

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6 0
1 year ago
Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
Deffense [45]

Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

r = \dfrac{a}{2\sqrt{2}}

where;

a = edge length.

Making "a" the subject of the formula:

a = 2 \sqrt{2} \times r

a = 2 \times 1.414 \times 135 \ pm

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

Density  \ of  \ unit \  cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}

Density of unit cell ( rhodium) = 12.279 g/cm³

3 0
2 years ago
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