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lidiya [134]
3 years ago
8

When hydrocarbon fuels burn in a plentiful supply of air they undergo complete combustion, forming carbon dioxide and water vapo

ur. If the air supply is limited incomplete combustion occurs and carbon monoxide and carbon may be formed. Describe the problems that can be caused by these products of complete and incomplete combustion.
Chemistry
1 answer:
Ksivusya [100]3 years ago
8 0
Problems of complete combustion. 1) CO2 is released into the environment, this is a greenshouse gas and contributes to global warming. 
Problems of incomplete combustion. 1) Carbon monoxide is toxic, as it's colourless, tasteless and odorless people cannot detect whether it is around them. 2) Carbon is formed in tiny pieces called particulates, these reach the atmosphere and cause global dimming. It can also provoke people with illnesses such as asthma.
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If a solution is strongly acidic, what<br> anionscannot be present in<br> appreciableconcentrations?
masha68 [24]

Explanation:

OH^- cannot be present in appreciable concentrations in a solution that is strongly acidic in nature. A weak acid is a complex that consists of hydrogen bound to an anion that does not dissociate well in solution comparing to a strong acid which dissolves completely to an anion. Therefore, OH^- cannot be present in appreciable concentrations in a solution that is strongly acidic in nature.  

7 0
3 years ago
Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
Firlakuza [10]

Answer:

  • [Ag⁺] = 1.3 × 10⁻⁵M
  • s = 3.3 × 10⁻¹⁰ M
  • Because the common ion effect.

Explanation:

<u></u>

<u>1. Value of [Ag⁺]  in a saturated solution of AgCl in distilled water.</u>

The value of [Ag⁺]  in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                    0

C          -s                      +s                  +s

E         X - s                   s                     s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × s
  • Ksp = s²

By substitution:

  • 1.8 × 10⁻¹⁰ = s²
  • s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

  • [Ag⁺] = 1.3 × 10⁻⁵M ← answer

<u></u>

<u>2. Molar solubility of AgCl(s) in seawater</u>

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                  0.54

C          -s                      +s                  +s

E         X - s                   s                     s + 0.54

The new equation for the Ksp equation will be:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × ( s + 0.54)
  • Ksp = s² + 0.54s

By substitution:

  • 1.8 × 10⁻¹⁰ = s² + 0.54s
  • s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

     

     s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

<u>3. Why is AgCl(s) less soluble in seawater than in distilled water.</u>

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

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Mechanical Advantage may be calculated using all of the following except ___.
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The only exception of calculating mechanical advantage is by multiplying resistance force by effort force. For example, in calculating the mechanical advantage of a lever, we consider its output and input forces. The equation would now be as follows:

Mechanical advantage = output force / input force
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Roman55 [17]

Answer:

What grade are you in. If i am older, I dont think you would be able to help me.

Explanation:

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Why and how do ions form?
Zanzabum

Answer and Explanation:

Ions are electrically charged particles that are formed from the removing and addition of electrons. It can be a positively or negatively charged atom.

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