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patriot [66]
3 years ago
6

Pb(OH)2 + _HCI → __ H2O + _PbCl2​

Chemistry
1 answer:
olganol [36]3 years ago
6 0
Pb(OH)2 + 2 HCl = 2 H2O + PbCl2
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According to the law of universal gravitation, any two objects are attracted to each other. The strength of the gravitational fo
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How are the molecular mass and molar mass of a compound similar and how are they different?
4vir4ik [10]

Answer:

Similarities: both state the mass of chemical species and they have the same numerical value

Differences: molecular mass refers to one single molecule and molar mass refers to one mole of a molecule

Explanation:

The molecular mass is the value of the mass of each molecule and it is measured in mass units (u). It is calculated adding the mass of each atom of the molecule.

The molar mass is the value of the mass of one mole of molecules, which means the mass of 6.022140857 × 10²³ molecules. The unit is g/mol.

For example, we can consider the methane molecule, which has the chemical formula of CH₄:

Molecular mass CH₄ = C mass + 4 x (H mass)

Molecular mass CH₄ = 12.01 + 4 x (1.01)

Molecular mass CH₄ = 16.05 u

Now to calculate the molar mass we multiply the value of the molecular mass by the Avogadro number and convert the units to g/mol:

Molar mass CH₄: 16.05 x \frac{1}{6.022140857 x 10x^{23} } g x 6.022140857 × 10²³ mol⁻¹

Molecular mass CH₄ = 16.05 g / mol

5 0
3 years ago
Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M
Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

  • 2AgNO₃(aq) + K₂CrO₄(aq)  → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

  • 0.20 M * 50.0 mL = 10 mmol of AgNO₃
  • 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

  • 4 mmol K₂CrO₄ * \frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4} = 1326.92 mg Ag₂CrO₄
  • 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
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