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4 years ago

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A solution has a [Ag+(aq)] of 0.01 M. The chloride concentration in solution is 1 x 10-5 M. Based on the following reaction, ans

wer the following questions: AgCl(s) Ag+(aq) + Cl-(aq), Ksp = 1.7 x 10-10 Calculate the value of the current solubility product. Is the K value obtained greater or less than Ksp? Will a precipitate of AgCl form?

1 answer:

earnstyle [38]4 years ago

7 0

The value of current solubility product is calculated as below

K = (Ag+)( Cl-)

Ag+ = 0.01 M
Cl-= 1 x10^-5M

K is therefore = 1 x10^-5 x 0.01 = 1 x10 ^ -7 M

The K obtained is greater than Ksp

that is K> KSp
1x10^-7 > 1.7 x10 ^-10

will precipitation of AgCl form?

yes the precipitation of AgCl will be formed since K> KSP

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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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3 years ago

What forces hold network solids together?

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Answer:

Covalent solids, also called network solids, are solids that are held together by covalent bonds. As such, they need localized electrons (shared between the atoms) and therefore the atoms are arranged in fixed geometries. Distortion far from this geometry can only occur through a breaking of covalent sigma bonds.

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4 years ago

How many liters of a 2.0 M solution of HCl do you need to have 8.0 moles of HCI?

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Answer:

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If an iron atom loses two electrons, the name of the iron ion is

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If iron atom looses 3 electrons, it forms Fe+3 (ferric) and if iron looses 2 electrons, it forms Fe+2(ferrous). Hope this helps!

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3 years ago

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8. What is the % weight of Nickel in Nickel Sulfamate (Ni(SO3NH2)2) ?

uranmaximum [27]

Answer:

% weight of nickle = 24 %

Explanation:

molar mass of Nickel Sulfamate (Ni(SO₃NH₂)₂) = 250.87 g/mol

Solution

1st we write down the molar mass of Ni

molar mass of Ni = 59 g/mol

now we write down the number of moles of Ni in (Ni(SO₃NH₂)₂)

number of moles of Ni = 1 mol

Now we calculate the mass of nickle present in (Ni(SO₃NH₂)₂)

<em> mass = moles × molar mass</em>

mass = 1 mol × 59 g/mol

mass = 59 g

now we calculate the % weight of nickle in (Ni(SO₃NH₂)₂)

<em> % weight = (weight of element ÷ total weight) × 100</em>

% weight of nickle = (59 ÷ 250.87) × 100

% weight of nickle = 0.24 × 100

% weight of nickle = 24 %

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3 years ago