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3 years ago

Consider the equilibrium

Chemistry

1 answer:

Kamila [148]3 years ago

7 0

Answer:

The answers are in the explanation

Explanation:

For the equilibrium:

B(aq) + H₂O(l) ⇌ HB⁺(aq) + OH⁻(aq).

By LeChatelier's principle, the increase in the concentration of a reactant (for example) at equilibrium will change the system counteracting the increasing producing more product.

Thus,

A) Will the equilibrium constant for the reaction increase, decrease, or stay the same? Why? .

The equilibrium constant is a thermodynamic constant that stay the same with the addition of a compound.

B) Will the concentration of HB⁺(aq) increase, decrease, or stay the same? Why?

By LeChatelier's principle, the addition of B will induce the formation of more HB⁺(aq) increasing the concentration.

C) Will the pH of the solution increase, decrease, or stay the same? Why?

As the addition of B induce the increasing of OH⁻, the pH of the solution will increase.

I hope it helps!

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For a representative element, the row number on the periodic chart will be the same as the shell number. TRUE OR FALSE

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What is the gram-formula mass of Fe(NO3)3?

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What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water?

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Answer:

The concentration of KBr is $C = 0.07899 \ mol L^{-1}$

Explanation:

From the question we are told that

The mass of KBr is $m_{KBr} = 2.35 \ g$

The molar mass of KBr is $M_{KBr} = 119 g/mol$

Volume of water is $V = 250 \ mL = 250 *10^{-3} = 0.250 \ L$

This implies that the volume of the solution is $V = 250 mL$

The number of moles of KBr is

$n = \frac{m_{KBr}}{M_{KBr}}$

Substituting values

$n = \frac{2.35}{119}$

$n = 0.01975 \ mol$

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$C = \frac{0.01975}{0.250}$

$C = 0.07899 \ mol L^{-1}$

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PLEASE HELP ME SOLVE THIS.Thank you so much!

Tju [1.3M]

Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

$Cr_{2}O^{2-}_{7} + Cl^{-} \rightarrow Cr^{3+} + Cl_{2}$

Now, the two half-reactions can be written as follows.

Reduction half-reaction: $Cr_{2}O^{2-}_{7} + 3e^{-} \rightarrow Cr^{3+}$

This will be balanced as follows.

$Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$ ... (1)

Oxidation half-reaction: $Cl^{-} \rightarrow Cl_{2} + 1e^{-}$

This will be balanced as follows.

$6Cl^{-} \rightarrow 3Cl_{2} + 6e^{-}$ ... (2)

Adding both equation (1) and (2) we will get the resulting equation as follows.

$Cr_{2}O^{2-}_{7} + 14H^{+} + 6Cl^{-} \rightarrow 2Cr^{3+} + 3Cl_{2} + 7H_{2}O$

Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.

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