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Marizza181 [45]
3 years ago
6

The Gateway Arch in St. Louis, MO is approximately 630 ft tall. How many U.S. half dollars would be in a stack of the same heigh

t? Each half dollar is 2.15 mm thick.
Chemistry
1 answer:
Paraphin [41]3 years ago
6 0

Answer:

The number of half dollars in the stack of the height of the Gateway Arch in St. Louis is 89,313 half dollars

Explanation:

The given height of the Gateway Arch in St. Louis = 630 ft

The thickness of each half dollar = 2.15 mm

By conversion of factors, we have;

1 feet = 304.8 mm

Therefore;

630 ft. = 304.8 × 630 mm = 192,024 mm

The number of half dollars that will stack up to 630 ft or 192,024 mm, is therefore, given as follows;

The number, n, of half dollars in the stack of the height of the Gateway Arch in St. Louis = The height of the Gateway Arch in St. Louis/(The thickness of each half dollar)

n = 630 ft/(2.15 mm/(half dollar)) = 192,024 mm/(2.15 mm/(half dollar)) = 89,313.49 ≈ 89,313 half dollars

The number of half dollars in the stack of the height of the Gateway Arch in St. Louis = 89,313 half dollars.

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Mass of CaCl₂ = 0.732 g

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

\tt \%(m/v)\rightarrow 12\%=\dfrac{mass~CaCl_2}{volume~of~solution}\times 100\%\\\\mass~CaCl_2=12\%\times 6.1\div 100\%\\\\mass~CaCl_2=0.732~g

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What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?
natita [175]

This question provides us –

  • Weight of \bf  KCl is = 47 g
  • Volume, V = 375 mL

__________________________________________

  • Molar Mass of \bf   KCl –

\qquad \twoheadrightarrow\bf  39.0983 \times 35.453

\qquad \twoheadrightarrow\bf 74.5513

<u>Using formula</u> –

\qquad \purple{\twoheadrightarrow\bf Molarity _{(Solution)} =  \dfrac{ W\times 1000}{MV}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{ 47 \times 1000}{74.5513\times 375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{47000}{27956.7375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \cancel{\dfrac{47000}{27956.7375}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = 1.68117M

\qquad \pink{\twoheadrightarrow\bf Molarity _{(Solution)}  = 1.7M}

  • Henceforth, Molarity of the solution is = 1.7M

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Write the net ionic equation for the reaction of aqueous solutions of ammonium chloride and iron(III) hydroxide.
vagabundo [1.1K]

Answer:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

Explanation:

Hello.

In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:

3NH_4Cl(aq)+Fe(OH)_3(s)\rightarrow 3NH_4OH+FeCl_3

And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:

3NH_4^+(aq)+3Cl^-(aq)+Fe(OH)_3(s)\rightarrow 3NH_4^+(aq)+3OH^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)

Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:

Fe(OH)_3(s)\rightarrow Fe^{3+}(aq)+3OH^-(aq)

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