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DedPeter [7]
3 years ago
7

If the balance were not tared prior to weighing out the KHP.... how would you expect this to affect the molarity of NaOH calcula

ted? What type of error is this?
Chemistry
1 answer:
aleksklad [387]3 years ago
6 0

Answer:

Following are the response to the given question:

Explanation:

In the given scenario, When the balance has never been tainted before the KHP is weighted, which can affect the molar concentration of NaOH because its molarity is directly proportional to the weight including its substance. In this question it is the mistake is systemic because it may be corrected by modifying balancing parameters.

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Arrange the procedural steps, from start to finish, that are required to prepare indigo from nitrobenzaldehyde and acetone in ba
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Acetone has α-hydrogens (on both sides) and thus can be deprotonated to give a nucleophilic enolate anion. The aldehyde carbonyl is much more electrophilic than that of a ketone, and therefore reacts rapidly with the enolate.

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Liquids and solids are referred to as "condensed phases" because the attractive forces cause molecules to become stationary. lea
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leave little space between the molecules.

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7 0
3 years ago
Read 2 more answers
At 700 K, the reaction 2SO2(g) + O2(g) &lt;====&gt; 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f
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Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

  • If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
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In this case:

Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }

Q=\frac{10^{2} }{0.10^{2} *0.10}

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

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