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3 years ago

Balance the equation and write the reaction-quotient expression, qc. u(s) + f2(g) ----> uf6(g)

Chemistry

2 answers:

nevsk [136]3 years ago

7 0

The balanced equation with the reaction-quotient expression is;

U(s) + 3F₂(g) → UF₆(g)

<h2>Further Explanation:</h2><h3>Chemical equations </h3>

Chemical equations are equations showing reactions between reactants to form products.
Chemical equations show the reactants or starting substances and products or substances formed during the reaction.

<h3>Law of conservation of mass </h3>

The law of conservation requires that when writing chemical equations, the mass of the reactants should be equal; to the mass of the products.
This is done by making sure the number of atoms of each element involved in the chemical equation is equal on both sides of the equation.

To ensure the law of conservation in chemical equations is observed we balance chemical equations.

<h3>Balancing chemical equation:</h3>

Balancing chemical equations is a try and error method that ensures the number of atoms in the side of the reactants is equal to the number of atoms in the side of products.
Balancing chemical equations may also require inclusion of state symbols which shows the state of each compound or element involved in the chemical reactions.
For example; the equation; U(s) + 3F₂(g) → UF₆(g) is balanced as the number of Uranium and Fluoride atoms are equal on both sides of the equation.

Keywords: Chemical equations, balancing of chemical equations

<h3>Learn more about:</h3>

Chemical equations: brainly.com/question/1566156
Balancing of chemical equations: brainly.com/question/1566156
Law of conservation of mass: brainly.com/question/1566156

Level: high school

Subject: Chemistry

Topic: Chemical equations

Sub-topic: Balancing chemical equations

Alona [7]3 years ago

4 0

The equation presented above is that of uranium reacting with fluorine forming uranium fluoride.

The chemical reaction can be balanced by carefully studying the equation and balancing the number of atoms of each of the element in both sides of the chemical reaction. That is,

<em> U(s) + 3F₂(g) --> UF₆(g)</em>

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If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of

Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.1\ g}{186.04\ g/mol}$

$Moles\ of\ ferrocene= 0.0005375\ mol$

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

$\rho=\frac{Mass}{Volume}$

or,

$Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g$

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.044\ g}{78.49\ g/mol}$

$Moles\ of\ acetyl\ chloride= 0.0005606\ mol$

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms 0.0005375 mole of monoacetylferrocene

Moles of product formed = 0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

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A dilute solution is a solution that has not reached the maximum concentration of solute dissolved in a solvent. The additional solute will dissolve when added in a dilute solution and will not appear in the aqueous phase. It is considered a state of dynamic equilibrium where the speeds in which the solvent dissolves the solute are greater than the recrystallization rate.

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