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Ivan
2 years ago
6

Describe the electron sharing between 2 oxygen atoms

Chemistry
1 answer:
erastova [34]2 years ago
3 0
When two oxygen atoms get close to each other, the attractions from the nucleus of both atoms attract the outer electrons.

(BRAINLIEST PLEASE!!!)
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Which of the following statements is true? A) This reaction will be spontaneous only at high temperatures. B) This reaction will
viktelen [127]

Answer:

D) This reaction will be nonspontaneous only at high temperatures.

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

  • ΔG < 0 , the reaction is Spontaneous
  • ΔG > 0 , the reaction is non Spontaneous
  • ΔG = 0 , the reaction is at equilibrium

The reaction has the value for ∆H = negative , and ∆S = negative ,

Now ,

∆G = ∆H -T∆S

     = ( - ∆H ) - T( - ∆S )

     =  ( - ∆H ) +T(  ∆S )

Now, for making the reaction Spontaneous ΔG = negative ,

Hence ,

The temperature is low, then the value for  ΔG will be negative , i.e. , Spontaneous reaction .

And , vice versa , at higher temperature , the reaction will have ΔG positive , and the reaction will be non -Spontaneous reaction .

The standard free energy of formation will be zero , only for the compounds that are in their pure form ,

Hence , Al(s) will have ΔG = 0 .

7 0
3 years ago
Can a combination reaction be a redox reaction ?
saul85 [17]

Answer:

Yes A redox reaction can be a combination reaction.

Two elements are mixed in a combination reaction to produce a single product.

Explanation:

Example: water formula

2H2 + O2 → 2H2O

Oxygen is reduced in this reaction when electrons are transferred from hydrogen to oxygen and hydrogen is also oxidized since oxygen from hydrogen accepts electrons. Oxygen is the oxidizing agent and hydrogen is the reducing agent.

2H2 + O2 → 2H2O

4 0
3 years ago
If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the
Sedbober [7]

Answer:

Let me give it a try.

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O

Balancing this reaction

2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

Moles= Molarity x Volume

Volume = 38.5ml = 0.0385L

Moles of Ca hydroxide = 0.150m/L x 0.0385L

(Notice the units canceling out...leaving moles).

=0.005775moles of Ca(OH)2.

From balanced reaction...

3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

0.005775moles of Ca(OH)2 would completely react with....

= 0.005775 x 2/(3)

=0.00385moles of H3PO4.

Now we're looking for its Concentration in Mol/L

Molarity=Moles of solute/Volume of solution(in L)

Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

Molarity = 0.00385/0.0535

=0.072Mol/L.

If this is wrong

then Simply Try The formula for Mixing of solutions

C1V1 = C2V2

0.15 x 38.5 = C2 x (15+38.5)

C2 = 0.11M/L.

7 0
2 years ago
Think about the decomposition reaction of ozone.
____ [38]

Answer:

3441.6 grams.

Explanation:

2 O₃ → 3 O₂

First we <u>convert 71.7 moles of ozone (O₃) into moles of oxygen gas (O₂)</u>, using the <em>stoichiometric coefficients</em> of the reation:

71.7 mol O₃ * \frac{3molO_2}{2molO_3} = 107.55 mol O₂

Then we <u>convert 107.55 moles of O₂ into grams</u>, using the <em>molar mass of O₂</em>:

107.55 mol * 32 g/mol = 3441.6 g

3 0
3 years ago
The table shows the amount of radioactive element remaining in a sample over a period of time.
Lunna [17]

Answer:

1. The half-life is 22 years.

2. 132 years

Explanation:

1. Determination of the the half-life.

The half-life of an element is the time taken for half the element to decay.

From the table given above, the original amount of the element 45 g. If we divide 45 by 2, we'll have 22.5 g as half the original amount of element.

Now, the time taken to obtain 22.5 g as shown from the table is 22 years.

Thus, the half-life the element is 22 years.

2. Determination of the time.

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Time (t) =?

Next, we shall the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

4.8125 = 1/2ⁿ × 308

Cross multiply

4.8125 × 2ⁿ = 308

Divide both side by 4.8125

2ⁿ = 308 / 4.8125

2ⁿ = 64

Express 64 in index form with 2 as the base.

2ⁿ = 2⁶

n = 6

Thus, 6 half-lives has elapsed.

Finally, we shall determine the time. This can be obtained as follow:

Number of half-lives (n) = 6

Half-life (t½) = 22 years

Time (t) =?

n = t / t½

6= t / 22 years

Cross multiply

t = 6 ×22

t = 132 years.

Thus, the time taken is 132 years.

6 0
3 years ago
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