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1 year ago

a mixture consisting of only of lithium chloride, lithium carbonate, and lithium nitrate was analyzed

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6 0

A mixture consisting of only of lithium chloride, lithium carbonate, and lithium nitrate was analyzed by the help of concentric hydrochloric acid.

<h3>What is concentric hydrochloric acid?</h3>

The concentric hydrochloric acid is used for the analysis of organic and inorganic mixture like lithium chloride, lithium carbonate, and lithium nitrate it releases fumes and cannot be touched with normal hands.

Always use a dropper to use it and for the chemical analysis.

Therefore, mixture consisting of only of lithium chloride, lithium carbonate, and lithium nitrate was analyzed by the help of concentric hydrochloric acid.

Learn more about concentric hydrochloric acid, here,

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Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

2 years ago

According to Bohr atomic model

nata0808 [166]

Answer:

A small positively charged nucleus surrounded by revolving negatively charged electrons in fixed orbits

7 0

2 years ago

A scientist has two samples of gas: The first sample contains one mole of argon atoms and has a mass of 39.948 g; the second sam

V125BC [204]

There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.

Explanation:

39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.

4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.

We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.

So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.

Learn more about:

Avogadro's number

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4 0

3 years ago

A 7.27-gram sample of a compound is dissolved in 250. grams of benzene. The freezing point of this solution is 1.02°C below that

UkoKoshka [18]

Answer:

The correct answer is 146 g/mol

Explanation:

<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:

ΔTf = Kf x m

Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:

ΔTf = 1.02ºC

Kf = 5.12ºC/m

From this, we can calculate the molality:

m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m

The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:

0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute

There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:

molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol

<em>Therefore, the molar mass of the compound is 146 g/mol </em>

6 0

2 years ago

Write a net ionic equation to show that benzoic acid, C6H5COOH, behaves as a Brønsted-Lowry acid in water.

Nitella [24]

Answer:

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

Explanation:

According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺. Let's consider the molecular equation showing that benzoic acid is a Brönsted-Lowry acid.

C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The complete ionic equation includes all the ions and molecular species.

C₆H₅COO⁻(aq) + H⁺(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)

5 0

2 years ago