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hram777 [196]
3 years ago
6

Osmium metal, the densest element, has a density of 22.6 g/mL, while hydrogen, the least dense element, has a density of 8.99 e-

5 g/mL. Calculate the volume occupied by 1.00 g of each element. How many times more dense is osmium than hydrogen?
Chemistry
1 answer:
KengaRu [80]3 years ago
5 0

Answer:

1. Volume of osmium = 0.044 mL

2. Volume of Hydrogen = 11123.47 mL

3. Osmium is 251390 times denser than hydrogen

Explanation:

From the question given above, the following data were obtained:

Density of osmium = 22.6 g/mL

Density of Hydrogen = 8.99×10¯⁵ g/mL

Mass of osmium = 1 g

Mass of Hydrogen = 1 g

1. Determination of the Volume of osmium.

Density of osmium = 22.6 g/mL

Mass of osmium = 1 g

Volume of osmium =?

Density = mass /volume

22.6 = 1 / volume

Cross multiply

22.6 × volume = 1

Divide both side by 22.6

Volume = 1 / 22.6

Volume of osmium = 0.044 mL

2. Determination of the Volume of Hydrogen.

Density of osmium = 8.99×10¯⁵ g/mL

Mass of osmium = 1 g

Volume of osmium =?

Density = mass /volume

8.99×10¯⁵ = 1 / volume

Cross multiply

8.99×10¯⁵ × volume = 1

Divide both side by 8.99×10¯⁵

Volume = 1 / 8.99×10¯⁵

Volume of Hydrogen = 11123.47 mL

3. Determination of the number of times osmium is denser than hydrogen.

Density of osmium (Dₒ) = 22.6 g/mL

Density of Hydrogen (Dₕ) = 8.99×10¯⁵ g/mL

Dₒ / Dₕ = 22.6 / 8.99×10¯⁵

Dₒ / Dₕ = 251390

Cross multiply

Dₒ = 251390 × Dₕ

Thus, osmium is 251390 times denser than hydrogen.

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Concentration of unknown acid is 0.061 M

Given:

Concentration of NaOH = 0.125 M

Volume of NaOH = 24.68 mL

Volume of acid solution = 50.00 mL

To Find:

Concentration of the unknown acid

Solution: Concentration is the abundance of a constituent divided by the total volume of a mixture. The concentration of the solution tells you how much solute has been dissolved in the solvent

Here we will use the formula for concentration:

M1V1 = M2V2

0.125 x 24.68 = 50 x M2

M2 = 0.125 x 24.68 / 50

M2 = 0.061 M

Hence, the concentration of unknown acid is 0.061 M

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3 years ago
When an object is lifted 6 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 18
AlexFokin [52]

Answer:

The answer to your question is the letter C. three times as much

Explanation:

Data

First step = 6 m

Second step = 18 m

Potential energy is the energy stored that depends on its position.

Formula

Pe = mgh

m = mass; g = gravity; h = height

Potential energy of the first step

        Pe1 = 6mg

Potential energy of the second step

        Pe2 = 18mg

-Divide the Pe2 by the Pe1

        Pe2/Pe1 = 18mg/6mg

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7 0
3 years ago
In the reaction of 675.9 grams of barium chloride and excess silver(I) nitrate, how many grams of silver(I) chloride should get
zvonat [6]

Taking into account the reaction stoichiometry, the mass of silver(I) chloride formed is 930.37 grams.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + BaCl₂ → 2 AgCl + Ba(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • AgNO₃: 2 moles
  • BaCl₂: 1 mole
  • AgCl: 2 moles
  • Ba(NO₃)₂: 1 mole

The molar mass of the compounds is:

  • AgNO₃: 169.87 g/mole
  • BaCl₂: 208.24 g/mole
  • AgCl: 143.32 g/mole
  • Ba(NO₃)₂: 261.34 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

AgNO₃: 2 moles ×169.87 g/mole= 339.74 grams

BaCl₂: 1 mole ×208.24 g/mole= 208.24 grams

AgCl: 2 moles ×143.32 g/mole= 286.64 grams

Ba(NO₃)₂: 1 mole×261.34 g/mole= 261.34 grams

<h3>Mass of silver(I) chloride formed</h3>

The following rule of three can be applied:  if by reaction stoichiometry 208.24 grams of barium chloride form 286.64 grams of silver(I) chloride, 675.9 grams of barium chloride form how much mass of silver(I) chloride?

mass of silver(I) chloride=\frac{675.9 grams of barium chloride x286.64 grams of silver(I) chloride }{208.24 grams of barium chloride}

mass of silver(I) chloride= 930.37 grams

Finally, the mass of silver(I) chloride formed is 930.37 grams.

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<em>An atom is called the mass number. </em><em />
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