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Tems11 [23]
4 years ago
5

____ HBr + ____ Mg(OH)2 ---> ____ MgBr2 + ____ H2O

Chemistry
1 answer:
swat324 years ago
4 0
Hello Camkirkland,
I think that you are trying to balance this equation.
In order to balance a chemical equation, the numbers of atoms of each element must be equal on both sides of the equation.

In this particular equation, the answer would be (2) HBr + (1) Mg(OH)2 ---> (1) MgBr2 + (2) H2O.

Hope this answers your question!
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Write a balanced equilibrium equation for the dissolution of nai in water. include phases.
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So the dissolution equation of NaI is:
NaI(s) → Na^+(s)  +   I^-(Aqu) 
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3 years ago
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The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C
soldi70 [24.7K]

Based on the information given regarding the energy, the true statement is C.  "The number of particles able to undergo a chemical reaction is less than the number that is not able to."

<h3>What is energy?</h3>

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In this case, the graph is shown with two inverted graph curves running close to each other, and one of the curves labeled T1 is slightly more spread out than the other labeled T2.

Here, T1 is the only sample with a particle able to react. Particle B is more likely to participate in the reaction than particle A and the number of particles able to undergo a chemical reaction is less than the number that is not able to.

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6 0
3 years ago
Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

The pH of the solution ;

pH=-\log[H^+]

=-\log[0.0001901 M]=3.72

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