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Sidana [21]
3 years ago
10

How many of the following statements about silver acetate, AgCH3COO, are true? i) More AgCH3CoO(S) will dissolve if the pH of th

e solution is reduced to 2.5 i) Less AgCH3CO0(s) will dissolve if AgNO; (s) is added to the aqueous solution iii) More AgCH3CoO(s) will dissolve if NaOH (aq) is added to the aqueous solution a) 0 b) c) 2 d) 3
Chemistry
1 answer:
shutvik [7]3 years ago
8 0

Answer:

All three statements are true

Explanation:

Solubility equilibrium of silver acetate:

AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}

  • If pH is increased then concentration of H^{+} increases in solution resulting removal of CH_{3}COO^{-} by forming CH_{3}COOH. Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more AgCH_{3}COO will dissolve
  • If AgNO_{3} is added then concentration of Ag^{+} increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less AgCH_{3}COO will dissolve
  • Insoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of Ag^{+}. So, more AgCH_{3}COO will dissolve

Hence all three statements are true

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Which structural formula does not represent benzene? a hexagon with nothing drawn inside it a hexagon with a dotted circle drawn
boyakko [2]
Answer:
            <span>A hexagon with nothing drawn inside it does not represent Benzene.

Explanation:

Below attached picture contains structures mentioned in given option.

Structure of Option-A is shown in Red Color.
</span>
Structure of Option-B is shown in Blue Color.

Structure of Option-C is shown in Green Color.

Structure of Option-D is shown in Orange Color.

Result:
          Except Option-A, all structures represent Benzene. Option-A in fact shows the structure of Cyclohexane.

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Do all multicellular organisms follow the same pattern of organization?
natta225 [31]
So so sorry if i'm wrong but i'm pretty sure no
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What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze
enyata [817]

Answer : 17.12 g

Explanation:\Delta T =k_b\times m

\Delta T = elevation in boiling point

k_b = boiling point elevation constant

m= molality

molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}

given \Delta T=6.28^{\circ}C

Molar mass of solute = 46.0 gmol^{-1}

Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

molality=\frac{x}{46gmol^{-1}\times0.15kg}

6.28 =2.53^{\circ}Ckgmol^{-1}\frac{x}{46gmol^{-1}\times 0.15kg}

x = 17.12 g



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3 years ago
26.8 moles of argon contain how many atoms?
Maru [420]
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3 years ago
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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

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3 years ago
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