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irinina [24]
3 years ago
9

Of the choices below, which is true for the relationship shown? It is Ka for the acid H3P2O72−. It is Kb for the acid H3P2O72−.

It is Ka for the acid H2P2O72−. It is Kb for the acid H2P2O72−.
Chemistry
2 answers:
Naily [24]3 years ago
3 0
The expression for the Ka for the given acid is:

Ka = [H2P2O7^2-] [H3O+] /[H3P2O7^2-]

<span>Ka is the acid dissociation constant or the acidity constant. It is a measure of the acid strength when in solution. It is an equilibrium constant for the dissociation of the acid.</span>
Anastasy [175]3 years ago
3 0

Answer:

A. It is Ka for the acid H3P2O72−

Explanation:

#platolivesmatter

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Please help I need the answer now. Write a balanced equation for combustion of pentane in plenty supply of air and in limited su
Ratling [72]

Pentane burns in plenty of air: CO₂ and H₂O is produced.

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Pentane burns in limited amount of air: CO or even C is produced along with H₂O.

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

<h3>Explanation</h3>

Pentane is a hydrocarbon. There are five carbon atoms in each of its molecule. Its molecular formula will be C₅H₁₂.

Hydrocarbon fuels burn to produce CO₂ when there's plenty of air.

? C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

  • Among all species in this reaction, C₅H₁₂ has the largest number of atoms per molecule. Assume that the coefficient of C₅H₁₂ is one.

<em>1</em> C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

  • C₅H₁₂ is the only <em>reactant</em> that contains C atoms. There are 5 C atoms in a C₅H₁₂ molecule. There should be the same number of C atoms in the products.
  • CO₂ is the only <em>product</em> that contains C atoms. There are one C atom in each CO₂ molecule. 5 C atoms correspond to 5 CO₂ molecules.

<em>1 </em>C₅H₁₂ + ? O₂ → <em>5</em> CO₂ + ? H₂O

  • Similarly, C₅H₁₂ is the only <em>reactant</em> that contains H atoms. H₂O is the only <em>product</em> that contains H atoms. There are 12 H atoms in one C₅H₁₂ molecule, which corresponds to 6 H₂O molecules.

<em>1</em> C₅H₁₂ + ? O₂ → <em>5</em> CO₂ + <em>6</em> H₂O

  • Both CO₂ and H₂O are <em>products</em> that contains O atoms. There are 5 × 2 + 6 × 1 = 16 O atoms in total in 5 CO₂ molecules and 6 H₂O molecules. The 16 O atoms on the <em>product</em> side corresponds to 8 O₂ molecules on the reactant side.

<em>1</em> C₅H₁₂ + <em>8</em> O₂ → <em>5</em> CO₂ + <em>6</em> H₂O

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

  • All coefficients shall be whole numbers. If there's any fraction in this equation, multiply both sides by the least common multiple of their denominators.

Hydrocarbon fuels burn to produce H₂O and CO when there's a limited supply of air. C (soot) might also be produced. Assuming that only CO is produced. Try to balance the equation using the same method.

1 C₅H₁₂ + 11/2 O₂ → 5 CO + 6 H₂O

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

Less O₂ is consumed for each mole of C₅H₁₂.

Consider: What would be the balanced equation when only C is produced?

<h3>Reference</h3>

"Products and effects of combustion", <em>GCSE Chemistry (Single Science)</em>, BBC Bitesize.

5 0
2 years ago
Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0
2 years ago
Charlie is balancing an equation. She has identified the atoms and counted the number of each in the reactants and products. Wha
Alecsey [184]
The answer to your question is Coefficients. 

*Hope this helps you! Feel free to message me if you have any further questions. Have a nice day! :)*
8 0
2 years ago
Read 2 more answers
Help someone!!!I will mark you brainliest
iVinArrow [24]
A) Na2S
b) AlF3
c) O2
d) C6H12O6
4 0
2 years ago
Read 2 more answers
Write the empirical formula for at least four ionic compounds that could be formed from the following ions:
CaHeK987 [17]

Answer:

Fe(CN)₂,  FeCO₃,  Pb(CN)₄,  Pb(CO₃)₂

Explanation:

Cations (positively charged ions) can only form ionic bonds with anions (negatively charged ions). However, you can't just simply put one cation and one anion together to form a compound. Each compound needs to been neutral, or have an overall charge of 0. When cations and anions do not have charges that perfectly cancel, you need to modify the amount of each ion in the compound.

1.) Fe(CN)₂
-----> Fe²⁺ and CN⁻

-----> +2 + (-1) + (-1) = 0

2.) FeCO₃

-----> Fe²⁺ and CO₃²⁻

-----> +2 + (-2) = 0

3.)  Pb(CN)₄

-----> Pb⁴⁺ and CN⁻

-----> +4 + (-1) + (-1) + (-1) + (-1) = 0

4.) Pb(CO₃)₂

-----> Pb⁴⁺ and CO₃²⁻

-----> +4 +(-2) + (-2) = 0

5 0
1 year ago
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