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3 years ago

7

The bond between metal ions and the sea of electrons

1 answer:

Vilka [71]3 years ago

6 0

Answer: Metallic Bonding. ... In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.

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Why is the krebs cycle also known as the citric acid cycle.

ololo11 [35]

Answer:

Krebs cycle is known as citric acid cycle because the citric acid is the first product formed in Krebs cycle. This cycle is also known as TCA cycle because the first product (citric acid) formed in Krebs cycle is a tricarboxylic acid (TCA).

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2 years ago

Suppose you have 11.0 mol of (CH4) and 9.0 mol of (O2) in a reactor. Calculate the largest amount of CO2 that could be produced

VARVARA [1.3K]

Hi just a reminder don’t click the links, and 11.0 is correct

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3 years ago

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sesenic [268]

Answer:

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Explanation:

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3 years ago

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The method used by Joseph Priestley to obtain oxygen made use of the thermal decomposition of mercuric oxide given below. What v

sukhopar [10]

Answer:

The volume of the oxygen gas is 0.246 L

Explanation:

Step 1: Data given

Temperature = 39 °C = 312 K

Temperature = 725 torr = 725 / 760 atm = 0.953947 atm

Mass of mercuric oxide = 3.97 grams

Molar mass of mercuric oxide = 216.59 g/mol

Step 2: The balanced equation

2HgO → 2Hg + O2

Step 3: Calculate moles mercuric oxide

Moles = mass / molar mass

Moles HgO = 3.97 grams / 216.59 g/mol

Moles HgO = 0.0183 moles

Step 3: Calculate moles oxyen

For 2 moles HgO we'll have 2 moles Hg and 1 mol O2

For 0.0183 moles HgO we'll have 0.0183/2 = 0.00915 moles O2

Step 4: Calculate volume O2

p*V = n*R*T

⇒with p = the pressure of the gas = 0.953947 atm

⇒with V = the volume of O2 gas = TO BE DETERMINED

⇒with n = the moles of O2 = 0.00915 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 312 K

V = (n*R*T)/p

V = (0.00915 moles * 0.08206 L*atm/mol*K * 312 K ) / 0.953947 atm

V = 0.246 L

The volume of the oxygen gas is 0.246 L

3 0

3 years ago

A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the

Kamila [148]

The answer for the following problem is mentioned below.

<u><em>Therefore the final temperature of the gas is 740 K</em></u>

Explanation:

Given:

Initial pressure of the gas ( $P_{1}$ ) = 1.8 atm

Final pressure of the gas ( $P_{2}$ ) = 4 atm

Initial temperature of the gas ( $T_{1}$ ) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas ( $T_{2}$ )

We know;

From the ideal gas equation;

we know;

P × V = n × R × T

So;

we can tell from the above equation;

<u> P ∝ T</u>

(i.e.)

<em> </em> $\frac{P}{T}$ <em> = constant</em>

$\frac{P_{1} }{P_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

$P_{1}$ = initial pressure of a gas

$P_{2}$ = final pressure of a gas

$T_{1}$ = initial temperature of a gas

$T_{2}$ = final temperature of a gas

$\frac{1.8}{4}$ = $\frac{333}{T_{2} }$

$T_{2}$ = $\frac{333*4}{1.8}$

$T_{2}$ = 740 K

<u><em>Therefore the final temperature of the gas is 740 K</em></u>

8 0

3 years ago