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Aleks04 [339]
4 years ago
8

The density of mercury is 13.6 g/mL what is the density in lbs/L ( 1 lb hint =0.454 kg )

Chemistry
1 answer:
Vesna [10]4 years ago
6 0

Answer:

30.0\frac{lb}{L}

Explanation:

Hello,

In this case, since 454 g are equivalent to 1 pound and 1000 millilitres are equivalent to 1 liter, the required density is computed below by applying the corresponding conversion factor:

=13.6\frac{g}{mL} *\frac{1lb}{454g} *\frac{1000mL}{1L} \\\\=30.0\frac{lb}{L}

Regards.

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(a)Determine the number of KNO3 molecules in 0.750 mol KNO3.
svp [43]

A. The number of molecules in 0.750 mole of KNO₃ is 4.515×10²³ molecules

B. The mass (in milligrams) of 2.39×10²⁰ molecules of Ag₂SO₄ is 124 mg

C. The number of molecules in 3.429 g of NaHCO₂ is 3.04×10²² molecules

<h3>Avogadro's hypothesis </h3>

1 mole of substance = 6.02×10²³ molecules

<h3>A. How to determine the number of molecules </h3>

1 mole of KNO₃ = 6.02×10²³ molecules

Therefore,

0.750 mole of KNO₃ = 0.75 × 6.02×10²³

0.750 mole of KNO₃ = 4.515×10²³ molecules

<h3>B. How to determine the mass of Ag₂SO₄</h3>

6.02×10²³ molecules = 312 g of Ag₂SO₄

Therefore,

2.39×10²⁰ molecules = (2.39×10²⁰ × 312) / 6.02×10²³

2.39×10²⁰ molecules = 0.124 g

Multiply by 1000 to express in mg

2.39×10²⁰ molecules = 0.124 g × 1000

2.39×10²⁰ molecules = 124 mg of Ag₂SO₄

<h3>C. How to determine the number of molecules </h3>

68 g of NaHCO₂ = 6.02×10²³ molecules

Therefore,

3.429 g of NaHCO₂ = (3.429 × 6.02×10²³) / 68

3.429 g of NaHCO₂ = 3.04×10²² molecules

Learn more about Avogadro's number:

brainly.com/question/26141731

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2 years ago
What is the shape around each carbon atom in ethylene, H C CH ? The represents a double bond.
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3 years ago
Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces
Brilliant_brown [7]

Answer:

0.364

Explanation:

Let's do an equilibrium chart for the reaction of combustion of ammonia:

2NH₃(g) + (3/2)O₂(g) ⇄ N₂(g) + 3H₂O(g)

4.8atm 1.9atm 0 0 Initial

-2x -(3/2)x +x +3x Reacts (stoichiometry is 2:3/2:1:3)

4.8-2x 1.9-(3/2)x x 3x Equilibrium

At equilibrium the velocity of formation of the products is equal to the velocity of the formation of the reactants, thus the partial pressures remain constant.

If pN₂ = 0.63 atm, x = 0.63 atm, thus, at equilibrium

pNH₃ = 4.8 - 2*0.63 = 3.54 atm

pO₂ = 1.9 -(3/2)*0.63 = 0.955 atm

pH₂O = 3*0.63 = 1.89 atm

The pressure equilibrium constant (Kp) is calculated with the partial pressure of the gases substances:

Kp = [(pN₂)*(pH₂O)³]/[(pNH₃)²*(pO_2)^{3/2}]

Kp = [0.63*(1.89)³]/[(3.54)²*(0.955)^{3/2}]

Kp = 4.2533/11.6953

Kp = 0.364

7 0
3 years ago
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