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2 years ago

Advantages and disadvantages of Law of Triad :

Chemistry

1 answer:

il63 [147K]2 years ago

7 0

Answer:

Dobereiner could identify only three triads. He was not able to prepare triads of all the known elements. All the known elements could not be arranged in the form of triads. For very low mass or for very high mass elements, the law was not holding good.

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Vikki [24]

The answer is ignore this thanks

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3 years ago

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Describe the Rutherford model... and and what are its limitation... going offline forever ✌️

SashulF [63]

Answer:

Limitations of Rutherford Atomic Model

Although the Rutherford atomic model was based on experimental observations it failed to explain certain things. Rutherford proposed that the electrons revolve around the nucleus in fixed paths called orbits. ... Ultimately the electrons would collapse in the nucleus.

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2 years ago

URGENT CHEMISTRY EXPERT!

vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na$_{3}$PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}$

Part 2. Net ionic equation

(a) Molecular equation

$\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})$

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

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3 years ago

The density of liquid mercury is 13.6 g/mL. What is its density in units of ? (2.54 cm = 1 in., 2.205 lb = 1 kg)

nalin [4]

Correct question

The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in3? (2.5 cm = 1 in., 2.205 lbs= 1 kg., 1000 g =1 kg, 1 mL = 1 cm³)

Answer:

$\rho0.4916\ lb/in^3$

Explanation:

Given that;-

The density = 13.6 g/mL

Also, 1 kg = 2.205 lb

1 kg = 1000 g

So, 1000 g = 2.205 lb

1 g = 0.002205 lb

Also,

1 in = 2.54 cm

1 in³ = 16.39 cm³

1 cm³ = 1 mL

So, 1 in³ = 16.39 mL

1 mL = 0.061 in³

The expression for the calculation of density is shown below as:-

$\rho=\frac{m}{V}$

Thus,

$\rho=\frac{13.6\ g}{1\ mL}=\frac{13.6\times 0.002205\ lb}{0.061\ in^3}=0.4916\ lb/in^3$

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3 years ago

Why is helium heavier than hydrogen

Kaylis [27]

Helium has more protons making it heavier by one atomic mass.

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3 years ago

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