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Nutka1998 [239]
3 years ago
14

The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in​3​? (2.5 cm+1 in., 2.205 lbs= 1 kg., 1000 g =1

kg, 1 mL = 1 cm3).
Chemistry
1 answer:
shusha [124]3 years ago
3 0

Answer:

Density, \rho=0.49\ lb/in^3

Explanation:

It is given that the density of liquid mercury is 13.6 g/mL. We need to convert the density into lb/in³.

We know that,

2.205 lbs= 1 kg

1 g = 0.0022 lb

1 mL = 0.0610 in³

13.6\ \dfrac{g}{mL}=13.6\times \dfrac{0.0022\ lb}{0.0610\ in^3}\\\\=0.49\ lb/in^3

So, the density of liquid mercury is 0.49\ lb/in^3.

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If a substance melts at 20oC, what is its freezing point?
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3 years ago
. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH
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The mass of ammonium chloride that must be added is : ( A ) 4.7 g

<u>Given data :</u>

Volume of water ( V )  = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI  +  H₂O  ⇄  NH₃ + H₃O⁺

where conc of H₃O⁺

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∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺  in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

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Determine the number of moles of NH₄CI in the solution

n = C . V

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Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

         = 0.08979 * 53.5 g/mol

         = 4.80 g  ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

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