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2 years ago

Balance this equation _Al +_HCI --> _H2 +_AICIz

Chemistry

1 answer:

kumpel [21]2 years ago

5 0

2Al+6HCl⇒3H₂+2AlCl₃

<h3>Further explanation </h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1

Reaction

Al+HCl⇒H₂+AlCl₃

Give coefficient

aAl+bHCl⇒cH₂+AlCl₃

Make equation

Al, left=a, right=1⇒a=1

Cl, left=b, right=3⇒b=3

H, left=b, right=2c⇒b=2c⇒3=2c⇒c=3/2

the equation becomes :

Al+3HCl⇒3/2H₂+AlCl₃ x2

2Al+6HCl⇒3H₂+2AlCl₃

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The progress of the reaction:

Blababa [14]

Answer:

The Equilibrium constant K is far greater than 1; K>>1

Explanation:

The equilibrium constant, K, for any given reaction at equilibrium, is defined as the ratio of the concentration of the products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their stoichiometric coefficients.

It tells us more about how how bigger or smaller the concentration of products is to that of the reactants when a reaction attains equilibrium. From the given data, as the color of the reactant mixture (Br2 is reddish-brown, and H2 is colourless) fades, more of the colorless product (HBr is colorless) is being formed as the reaction approaches equilibrium. This indicates yhat the concentration of products becomes relatively higher than that of the reactants as the reaction progresses towards equilibrium, the equilibrium constant K, must be greater than 1 therefore.

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3 years ago

How many moles are in 7.1x10^21 atoms of iron?

Nitella [24]

Answer:

0.011 moles

Explanation:

There are about 6.02*10^23 atoms in a mole, so in the given sample, there are

$\frac{7.01 \times {10}^{21} }{6.02 \times {10}^{23} }$

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8 0

2 years ago

Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying

PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? ( $K_a$ for HF is $6.8\times 10^{-4}$ .)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = $K_a=6.8\times 10^{-4}$

$HF\rightleftharpoons H^++F^-$

Initially

c 0 0

At equilibrium :

(c-x) x x

The expression of disassociation constant is given as:

$K_a=\frac{[H^+][F^-]}{[HF]}$

$K_a=\frac{x\times x}{(c-x)}$

$6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}$

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

$[H^+]=x=0.01346 M$

The pH of the solution is ;

$pH=-\log[H^+]=-\log[0.01346 M]=1.87$

The pH of an 0.280 M HF solution is 1.87.

6 0

2 years ago

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Lera25 [3.4K]

Answer:

D. 4

Explanation:

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