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3 years ago

Could someone explain what oxidation numbers are and what these questions are asking of me?

Chemistry

1 answer:

Oxana [17]3 years ago

6 0

The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation of an atom in a chemical compound.

Explanation:

The oxidation number of an atom is the charge that atom would have if the compound was composed of ions. 1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. The oxidation number of simple ions is equal to the charge on the ion.

The oxidation number of a mono atomic ion equals the charge of the ion. The oxidation number of H is +1, but it is -1 in when combined with less electro negative elements. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides. The oxidation number of a Group 1 element in a compound is +1.

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Predict the sign for AS for each of the following systems: Water freezing Water evaporating Crystalline urea dissolving Assembly

djyliett [7]

Answer: 1. Water freezing : $\Delta S$ is -ve.

2. Water evaporating : $\Delta S$ is +ve.

3. Crystalline urea dissolving : $\Delta S$ is +ve.

4. Assembly of the plasma membrane from individual lipids: $\Delta S$ is -ve.

5. Assembly of a protein from individual amino acids: $\Delta S$ is -ve.

Explanation:

Entropy is defined as the measurement of degree of randomness in a system.

It is represented by symbol S and we can only measure a change in entropy which is given by $\Delta S$ .

If there is decrease in randomness , the sign for $\Delta S$ is -ve and If there is increase in randomness , the sign for $\Delta S$ is +ve.

1. Water freezing: Entropy decreases as we move from liquid state to to solid state and thus $\Delta S$ is -ve.

2. Water evaporating : Entropy increases as we move from liquid state to gaseous state and thus $\Delta S$ is +ve.

3. Crystalline urea dissolving : The molecules convert from solid and ordered state to aqueous phase and random state. Thus the entropy increases and thus $\Delta S$ is +ve.

4. Assembly of the plasma membrane from individual lipids: random lipids are associating to form a single large polymer and thus entropy decreases and thus $\Delta S$ is -ve.

5. Assembly of a protein from individual amino acids: random amino acids are associating to form a single large polymer and thus entropy decreases and thus $\Delta S$ is -ve.

5 0

2 years ago

4.07 x 10-17 (5.6 x 10") (5.8 x 105)

Andrej [43]

4.07 x 10-17 = 23.7
(5.6 x 10”) (5.8 x 105)= log(100)
log(7)
Hope this helped

3 0

2 years ago

Summarize the weather conditions related to a cold front. Remember to include all data collected on cold fronts in this activity

svp [43]

Answer: Many fronts cause weather events such as rain, thunderstorms, gusty winds and A weather front is a transition zone between two different air masses at the Lifted warm air ahead of the front produces cumulus or cumulonimbus clouds

Explanation:

Hope this was helpful

5 0

2 years ago

Which of the following is NOT a derived unit? a) <img src="https://tex.z-dn.net/?f=cm%5E%7B3%7D" id="TexFormula1" title="

seraphim [82]

Answer:

c) kg

Explanation:

Kilograms stands alone. It has to be hooked up to another unit for it to be a derived unit.

I am joyous to assist you anytime.

4 0

2 years ago

Read 2 more answers

fills a 500.mL flask with 3.6atm of carbon monoxide gas and 1.2atm of water vapor. When the mixture has come to equilibrium she

enot [183]

Answer:

The answer to the question is

The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits

Explanation:

To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure

At the first trial the mixture contains

3.6 atm CO

1.2 atm H₂O (g)

Total pressure = 3.6+1.2= 4.8 atm

which gives

3.36 atm CO

0.96 atm H₂O (g)

0.24 atm H₂ (g)

That is

CO+H₂O→CO(g)+H₂ (g)

therefore the mixture contained

0.24 atm CO₂ and the total pressure =

3.36+0.96+0.24+0.24 = 4.8 atm

when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂

At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857

adding 1.8 atm CO gives 4.46 atm hence we have

(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857

which gives x = 0.031 atm or x = -0.6183 atm

Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm

7 0

3 years ago