Describe the contributions made by the following scientists to the development of the periodic table.

In which of these diatomic molecules would you NOT find an octet of electrons

Semmy [17]

Answer:

The hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

Explanation:

Let's evaluate each case.

1. Nitrogen (N₂):

With Z = 7, nitrogen has the following electronic configuration

1s²

2s² 2p³ → valence electrons

Since its valence electrons are 5, in the molecule one nitrogen atom shares 3 electrons with the other one, and each remains with an electron pair, so each atom has an octet of electrons.

2. Hydrogen (H₂):

With Z = 1, its electronic configuration is:

1s¹ → valence electron

In the molecule, the hydrogen atoms share the only electron they have, so they will have only 2 electrons around. In this diatomic molecule, we can not find an octet.

3. Oxygen (O₂):

Z = 8. Electronic configuration:

1s²

2s² 2p⁴ → valence electrons

In the diatomic molecule, each oxygen atom shares 2 electrons with the other one and remains with 2 pairs of electrons, therefore, each oxygen atom has an octet.

4. Fluorine (F₂)

Z = 9. Electronic configuration:

1s²

2s² 2p⁵ → valence electrons

In this molecule, each fluorine atom shares 1 electron with the other and remains with 3 pairs of electrons, hence, each fluorine atom has an octet of electrons around.

Finally, we can say that the hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

I hope it helps you!

8 0

3 years ago

A) Compute the repeat unit molecular weight of polystyrene. B) Compute the number-average molecular weight for a polystyrene for

Andrej [43]

Answer:

Explanation:

In Polystrene, the molecular formula for the repeat unit = $C_8H_8$ ;

and the atomic weights of Carbon C = 12.01 g/mol

For Hydrogen, it is 1.01 g/mol

Hence, the repeat unit molecular weight is:

m = 8 (12.01 g/mol)+8(1.01 g/mol)

m = 96.08 g/mol + 8.08 g/mol

m = 104.16 g/mol

The degree of polymerization = no-average molecular weight/repeat unit molecular weight.

Mathematically;

$DP = \dfrac{\overline M_n}{m}$

$\overline M_n= DP \times m$

$\overline M_n= 25000 \times 104.16 \ g/mol$

$\overline M_n= 2604000 \ g/mol$

7 0

3 years ago

What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperatu

tatyana61 [14]

Answer:

1) 6.0 atm.

2) 2.066 atm.

Explanation:

From the general law of ideal gases:

PV = nRT.

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

1) What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperature is held constant?

At constant T and at two different (P, and V):

P₁V₁ = P₂V₂.

P₁ = 2.0 atm, V₁ = 150.0 mL.

P₂ = ??? atm, V₂ = 50.0 mL.

∴ P₂ = P₁V₁/V₂ = (2.0 atm)(150.0 mL)/(50.0 mL) = 6.0 atm.

2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?

Since the container is rigid, so it has constant V.

At constant V and at two different (P, and T):

P₁/T₁ = P₂/T₂.

P₁ = 2.0 atm, T₁ = 30.0°C + 273 = 303 K.

P₂ = ??? atm, T₂ = 40.0°C + 273 = 313 K.

∴ P₂ = P₁T₂/T₁ = (2.0 atm)(313.0 K)/(303.0 K) = 2.066 atm.

3 0

4 years ago

A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

$T(K)=T(^{\circ}C)+273.15$

Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

6 0

4 years ago

Which type of bond is formed between atoms of solid gold

goldfiish [28.3K]

Answer:

covalent bonds

Explanation:

5 0

3 years ago

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