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3 years ago

Iron has a density of 7.86 g/cm3 and a mass of 3.76 kg. calculate the volume

1 answer:

7 0

Density = Mass / Volume

Input the values in their respective places and then use algebra to solve for volume.

Hope that helps.

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D. all of these can produce nuclear energy

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3 years ago

Draw a picture of the moon and sun illustrating the full moon phrase

jek_recluse [69]

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3 years ago

Read 2 more answers

What is largest value for the third quantum number

Kryger [21]

Answer:

18.

Explanation:

3s^2 3p^6 d^10

4 0

2 years ago

Iron iii chloride ammonium hydroxide balanced equation, complete ionic equation, net ionic equation

Vladimir [108]

Iron III Chloride has a chemical formula of FeCl₃, while ammonium hydroxide has a chemical formula of NH₄OH.

The balanced equation would be:

FeCl₃ (aq) + 3 NH₄OH (aq) → Fe(OH)₃ (s) + 3 NH₄Cl (aq)
The precipitate is Fe(OH)₃ or iron iii hydroxide.

To find the complete ionic equation, dissociate the compounds in aqueous phases into their ionic forms:

Fe³⁺ + Cl⁻ + NH₄⁺ + 3 OH⁻ --> Fe(OH)₃(s) + NH₄⁺ + Cl⁻

To find the net ionic equation, cancel out like ions that appear both in the reactant and product side:

Fe³⁺ + 3 OH⁻ --> Fe(OH)₃

4 0

3 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

3 years ago