All categories

3 years ago

In a piece of silver, where would you find the valence electrons?

Chemistry

2 answers:

poizon [28]3 years ago

6 0

Moving freely between the cations

deff fn [24]3 years ago

4 0

Answer: Option (A) is the correct answer.

Explanation:

The valence electrons are the electrons which are present in the outermost shell of an atom.

A silver atom contains a positive charge as silver is a cation.

In a piece of silver, the silver atoms will be linked together. As it is known that silver is a cation, it will hold a positive charge.

Therefore, it can be concluded that in a piece of silver, the valence electrons will be moving freely between the cations.

You might be interested in

Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− ,

zysi [14]

Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

CH₃CH₂CO₂H = HA

CH₃CH₂CO₂⁻ = A⁻

pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

(a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

$10^{-0.644}$ = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

(b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

$10^{-0.004}$ = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

(c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

$10^{0.426}$ = [A⁻]/[HA]

2.667 = [A⁻]/[HA]

6 0

3 years ago

What experimental evidence was provided for the nuclear model of an atom

UkoKoshka [18]

Alpha particles bouncing off of gold foil.

6 0

3 years ago

A student is titrating 50 mL of 0.32 M NH3 with 0.5 M HCl. How much hydrochloric acid must be added to react completely with the

Sveta_85 [38]

Answer:

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

Explanation:

Using the formula

Ca Va = Cb Vb

Cb = 0.32 M

Vb = 50 mL = 50/1000 = 0.050L

Ca = 0.5 M

Va =?

Substituting for Va in the equation, we obtain:

Va = Cb Vb / Ca

Va = 0.32 * 0.05 / 0.5

Va = 0.016 / 0.5

Va = 0.032 L

The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL

6 0

3 years ago

For which of the following elements would the most common ion be expected to have a larger radius than that of its corresponding

Andre45 [30]

Answer: The ion that is expected to have a larger radius than the corresponding atom is chlorine.

Explanation:

There are two types of ions:

Cations: They are formed when an atom looses its valence electrons. They are positive ions.
Anions: They are formed when an atom gain electrons in its outermost shell. They are negative ions.

For positive ions, the removal of electron increases the nuclear charge for an outermost electron because the outermost electrons are more strongly attracted by the nucleus. So, the effective nuclear charge increases for cations and thus, the size of the cation will be smaller than that of the corresponding atom.

For negative ions, the addition of electron decreases the nuclear charge for an outermost electron because the outermost electrons are less strongly attracted by the nucleus. So, the effective nuclear charge decreases for anions and thus, the size of the anion will be larger than that of the corresponding atom.

For the given options:

Option a: Chlorine

Chlorine gains 1 electron and form $Cl^-$ ion

Option b: Sodium

Sodium looses 1 electron and form $Na^+$ ion

Option c: Copper

Copper looses 2 electrons and form $Cu^{2+}$ ion

Option d: Strontium

Strontium looses 2 electrons and form $Sr^{2+}$ ion

Hence, the ion that is expected to have a larger radius than the corresponding atom is chlorine.

8 0

3 years ago

What type of energy has caused the fewest deaths per unit of energy produced?

mash [69]

Answer: Nuclear energy is by far the safest energy source in this comparison – it results in more than 442 times fewer deaths than the 'dirtiest' forms of coal; 330 times fewer than coal; 250 times less than oil; and 38 times fewer than gas.

4 0

3 years ago