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1 year ago

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If A is slowly added to a solution containing 0.0500 M of B and 0.0500 M of C, which solid will precipitate first? The solubilit

y product constant for A2B3 is 2.30×10−8. The solubility product constant for AC4 is 4.10×10−8.

1 answer:

OleMash [197]1 year ago

5 0

Answer:

AC₄ will precipitate out first.

Explanation:

A solid will precipitate out if the ionic product of the solution exceeds the solubility product.

Let us check the ionic product

a) A₂B₃

Ionic product = [A]²[B]³

[A] = say "s"

[B] = 0.05 , [B]³ = (0.05)³ = 0.000125

2.3 X 10⁻⁸ = [A]²(0.000125)

[A] = 0.0136

b) AC₄

Ionic product = [A] [C]⁴

[A] = "s"

[A][0.05]⁴ = 4.10 X 10⁻⁸

[A]=0.00656 M

So for ionic product to exceed solubility product, we need less concentration of A in case of AC₄.

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List the following compounds in decreasing electronegativity difference. cl2 hcl nacl

Mkey [24]

Based on Pauling Scale, electro negativity of Cl = 3.2, Na = 0.9 and H = 2.1

Thus, Electronegativity difference  in $Cl_{2}$ = 3.2 -3.2 = 0
Electronegativity difference  in NaCl = 3.2-0.9 = 2.3
Similarly, Electronegativity difference  in HCl = 3.2 - 2.1 = 1.1

Thus, among the listed molecules following is the decreasing order of electronegativity difference: NaCl> HCl > $Cl_{2}$

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1 year ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

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1 year ago

Methylene chloride (CH2Cl2) has fewer chlorine atoms than chloroform (CHCl3). Nevertheless, methylene chloride has a larger mole

icang [17]

Answer:

Explanation has been given below.

Explanation:

Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.

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2 years ago

Nh4i(aq)+koh(aq)→ express your answer as a chemical equation. identify all of the phases in your answer.

tester [92]

NH4I (aq) + KOH (aq) in chemical equation gives

NH4I (aq) + KOH (aq) = KI (aq) + H2O(l) + NH3 (l)

Ki is in aqueous state H2o is in liquid state while NH3 is in liquid state

from the equation above 1 mole of NH4I (aq) react with 1 mole of KOH(aq) to form 1mole of KI(aq) , 1mole of H2O(l) and 1 Mole of NH3(l)

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1 year ago

Read 2 more answers

Write the expression for the equilibrium constant Kp for the following reaction.Enclose pressures in parentheses and do NOT writ

maxonik [38]

<u>Answer:</u> The expression for $K_p$ is written below.

<u>Explanation:</u>

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_p$ is written as:

$K_p=\frac{P_{C}^c\times P_{D}^d}{P_{A}^a\times P_{B}^b}$

The partial pressure for solids and liquids are taken as 1.

For the given chemical equation:

$NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)$

The expression for $K_p$ for the following equation is:

$K_p=\frac{(P NH_3)\times (P H_2S)}{(P NH_4HS)}$

The partial pressure of $NH_4HS$ will be 1 because it is solid.

So, the expression for $K_p$ now becomes:

$K_p=\frac{(P NH_3)\times (P H_2S)}{1}$

Hence, the expression for $K_p$ is written above.

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2 years ago