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3 years ago

What is hydrolysis of a salt and give an example

Chemistry

1 answer:

GenaCL600 [577]3 years ago

6 0

Answer:

Dissolving a salt of a weak acid or base in water is an example of a hydrolysis reaction. Strong acids may also be hydrolyzed. For example, dissolving sulfuric acid in water yields hydronium and bisulfate.

Explanation:

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Given 450.98 g of Cu(NO3)2, how many moles of Ag can be made? Provide your final answer rounded to two decimal places.

vivado [14]

Answer:

4.82 moles of Ag.

Explanation:

We'll begin by calculating the number of mole in 450.98 g of Cu(NO₃)₂. This can be obtained as follow:

Molar mass of Cu(NO₃)₂ = 63.5 + 2[14 + (16×3)]

= 63.5 + 2[14 + 48]

= 63.5 + 2[62]

= 63.5 + 124

= 187.5 g/mol

Mass of Cu(NO₃)₂ = 450.98 g

Mole of Cu(NO₃)₂ =?

Mole = mass /Molar mass

Mole of Cu(NO₃)₂ = 450.98 / 187.5

Mole of Cu(NO₃)₂ = 2.41 moles

Next, we shall determine the number of mole of Cu needed to produce 450.98 g (i.e 2.41 moles) of Cu(NO₃)₂. This can be obtained as follow:

Cu + 2AgNO₃ —> Cu(NO₃)₂ + 2Ag

From the balanced equation above,

1 mole of Cu reacted to produce 1 mole of Cu(NO₃)₂.

Therefore, 2.41 moles of Cu will also react to produce 2.41 moles of Cu(NO₃)₂.

Thus, 2.41 moles of Cu is needed for the reaction.

Finally, we shall determine the number of mole of Ag produced from the reaction. This can be obtained as follow:

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of Ag.

Therefore, 2.41 moles of Cu will react to produce = 2× 2.41 = 4.82 moles of Ag.

Thus, 4.82 moles of Ag were obtained from the reaction.

6 0

3 years ago

Calculate the moles of 12.354 grams of c6h12o6

gavmur [86]

The correct answer is 0.06857 moles.

C₆H₁₂O₆, that is, glucose has six carbons, twelve hydrogens, and six oxygen atoms. The atomic weight of C, H and O are as follows:

Six atoms of carbon = 6 × 12.01 g = 72.06 g

Twelve atoms of hydrogen = 12 × 1.008 g = 12.096 g

Six atoms of oxygen = 6 × 16.00 g = 96.00 g

So, the molar mass of C₆H₁₂O₆ is 72.06 g + 12.096 g + 96.0 g = 180.156 g.

It can also be written in the form as 180.16 g of C₆H₁₂O₆ is equal to 1 mole of C₆H₁₂O₆or 180.16 g/mole (as the molar mass)

Now, there is a need to find moles of 12.354 grams of C₆H₁₂O₆. So, the final conversion is:

12.354 g C₆H₁₂O₆ × 1 mole of C₆H₁₂O₆ / 180.16 g C₆H₁₂O₆

= 0.06857 moles

5 0

3 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago

What are ionic compounds

ASHA 777 [7]

Answer: Ionic compounds are compounds consisting of ions.

Two-element compounds are usually ionic when one element is a metal and the other is a non-metal

Explanation: hope this helps!

7 0

3 years ago

you have been observing an insect that defends itself from enemies by secreting a caustic liquid. analysis of the liquid shows i

Alexus [3.1K]

pH of the buffer solution is 1.76.

Chemical dissociation of formic acid in the water:

HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)

The solution of formic acid and formate ions is a buffer.

[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions

[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate

[HCOOH] = 1.45 M - 0.015 M

[HCOOH] = 1.435 M; equilibrium concentration of formic acid

pKa = -logKa

pKa = -log 1.8×10⁻⁴ M

pKa = 3.74

Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)

pH = 3.74 + log (0.015 M/1.435 M)

pH = 3.74 - 1.98

pH = 1.76

More about buffer: brainly.com/question/4177791

#SPJ4

4 0

1 year ago

What is hydrolysis of a salt and give an example​

What is hydrolysis of a salt and give an example