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3 years ago

Describe the spin of carbon-14's valence electrons and explain how you came about your answer.

Chemistry

1 answer:

Snezhnost [94]3 years ago

8 0

we have to know the spin of valence electrons of carbon-14

There are four unpaired electron which are called as valence electron also.The spin of the four unpaired electron is either upfilled or down filled.

The ground state electronic configuration of C-atom is 1s²2s²2p² and one electron from 2s orbital gets excited to 2p orbital. The elctronic configuration in excited state is 1s²2s¹ $2p_{x}^{1} 2p_{y}^{1}2p_{z}^{1}$ .

The electron jumps because half-filled orbitals are more stable. Exchange energy is less than pairing energy.

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Answer:

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3 years ago

In atomic solids where the lattice points are occupied by Group 8 ( 18) atoms, the atoms are held together by __________ forces.

shtirl [24]

Answer:

London dispersion forces

Explanation:

There are different forces of attraction that helps to hold atoms or Molecules of a particular substance together. Some of the forces of attraction are ionic/ electrovalent bond, covalent bond, vander waals forces of attraction and so on.

Under the vander waals forces of attraction we have what is known as the London dispersion forces. This force of attraction is a very weak and it is commonly found in the atoms of noble gases.

The intermolecular force of attraction in which we are talking about that is london dispersion forces is formed as a result of the formation of non-polar dipoles which are not permanent.

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3 years ago

Part IV. Limiting Reactants! A Challenge Problem!

Alexxandr [17]

Answer:

a. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

b. Fe2O3 is the limiting reactant.

c. 6.30 grams Fe

d. 52.6 %

Explanation:

Step 1: Data given

Mass of iron(III) oxide Fe2O3 = 9.00 grams

Mass of aluminium = 4.00 grams

Molar mass Fe2O3 = 159.69 g/mol

Aluminium molar mass = 26.98 g/mol

Step 2: The balanced equation

Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

Step 3; Calculate Moles

Moles = mass / molar mass

Moles Fe2O3 = 9.00 grams / 159.69 g/mol

Moles Fe2O3 = 0.0564 moles

Moles Al = 4.00 grams / 26.98 g/mol

Moles Al = 0.148 moles

Step 4: Calculate limiting reactant

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

Fe2O3 is the limiting reactant. It will completely be consumed (0.0564 moles). Al is in excess. There will react 0.0564*2 = 0.1128 moles

There will remain 0.148 - 0.1128 = 0.0352 moles Al

Step 5: Calculate moles Fe

For 1 mol Fe2O3 we need 2 moles Al to produce 2 moles Fe and 1 mol Al2O3

For 0.0564 moles Fe2O3 we'll have 2*0.0564 = 0.1128 moles Fe

Step 6: Mass of Fe

Mass Fe = 0.1128 moles * 55.845 g/mol

Mass Fe = 6.30 grams

Step 7: If you carried out this reaction and it actually produced 0.475 mL of molten iron (r = 6.98 g/mL), what is the percent yield of this reaction?

Density = mass / volume

Mass = density * volume

Mass = 6.98 g/mL * 0.475 mL

Mass = 3.3155 grams

Percent yield = (actual mass / theoretical mass) * 100%

Percent yield = (3.3155 /6.30 ) * 100 %

Percent yield = 52.6 %

3 0

3 years ago

How many mL are in 7.3 dL?

Arte-miy333 [17]

Answer:

730 mL which is .73 L

Explanation:

8 0

3 years ago

Consider the half reactions below for a chemical reaction.

ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

7 0

3 years ago