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3 years ago

What is the pH of a solution with a concentration of 5.2 × 10–8 M H3O+?

Chemistry

1 answer:

lora16 [44]3 years ago

7 0

Answer : The pH of a solution is, 7.28

Solution : Given,

Concentration of hydronium ion, $H_3O^+$ = $5.2\times 10^{-8}M$

pH : It is defined as the negative logarithm of hydronium ion concentration or hydrogen ion concentration.

$pH=-\log [H_3O^+]$

Now put the value of hydronium ion concentration in this expression, we get the pH of the solution.

$pH=-\log (5.2\times 10^{-8})=7.28$

Therefore, the pH of a solution is, 7.28

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Which contain a common ion that will shift the equilibrium system represented by the equation shown? Select all that apply. MgSO

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The compounds in the reaction that shifts the equilibrium with the common ion effect are $\rm MgSO_2\;and\;HNO_3$ .

<h3>What is the common ion effect?</h3>

The equilibrium condition has an equal amount of reactant and products in the reaction.

The addition of ions same as product or reactant results in a change in the equilibrium of condition and shifts the equilibrium in opposite direction.

The given reaction is:

$\rm H_2SO_4\;\rightleftharpoons 2H^+\;+\;SO_4^{2-}$

The addition of sulfate and hydrogen ions shifts the equilibrium.

Thus, the compounds that shift the equilibrium for the reaction are $\rm MgSO_2\;and\;HNO_3$ .

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2 years ago

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Carbon is just a simple element where as water is a compound which means water is made up of 2 or more elements

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3 years ago

When ice is melting, the heat being added causes:

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Explanation:

Adding heat can cause ice (a solid) to melt to form water (a liquid). Removing heat causes water (a liquid) to freeze to form ice (a solid). When water changes to a solid or a gas, we say it changes to a different state of matter. Even though the water's physical form changes, its molecules stay the same.

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Be sure to answer all parts. Write the balanced equations corresponding to the following rate expressions: a) rate = − 1 3 Δ[CH4

Alinara [238K]

Answer : The balanced equations will be:

(a) $3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $2N_2O_5\rightarrow 2N_2+5O_2$

(c) $2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

(a) $Rate=-\frac{1}{3}\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=-\frac{d[CO_2]}{dt}=+\frac{1}{4}\frac{d[CH_3OH]}{dt}$

The balanced equations will be:

$3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $Rate=-\frac{1}{2}\frac{d[N_2O_5]}{dt}=+\frac{1}{2}\frac{d[N_2]}{dt}=+\frac{1}{5}\frac{d[O_2]}{dt}$

The balanced equations will be:

$2N_2O_5\rightarrow 2N_2+5O_2$

(c) $Rate=-\frac{1}{2}\frac{d[H_2]}{dt}=-\frac{1}{2}\frac{d[CO_2]}{dt}=-\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2CO_3]}{dt}$

The balanced equations will be:

$2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

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