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Svetllana [295]
3 years ago
6

Which colligative property is employed when salt is put on an icy sidewalk to melt ice?

Chemistry
2 answers:
Iteru [2.4K]3 years ago
6 0
<span>For this number, we are to determine the colligative property that is affected by the addition of salt in the sidewalk in order to melt the ice. The answer is that the colligative property being displayed is the freezing point depression. Colligative properties are innate properties of substances and are only dependent on the amount of substance being added. </span>
ankoles [38]3 years ago
4 0
<span>The correct answer is 'freezing point depression'. Colligative properties depend on the concentration of molecules of a solute. Examples of other colligative properties are boiling point elevation or vapour pressure lowering. The salt causes ice on the side walk to melt because it lowers the freezing point. </span>
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Which of these would most likely be one of the final steps when performing a strong acid-base titration? Prepare the burette. Re
DanielleElmas [232]

Answer is: Prepare to measure pH change.

For example for strong acid-base titration, sodium hydoxide and hydrochloric can be used.

Balanced chemical reaction: HCl + NaOH → NaCl + H₂O.

In this reaction pH of equivalence point will be always 7.

Equivalence point is the point which there is stoichiometrically equivalent amounts of acid and base.  

Chemist can draw pH curve (graph showing the change in pH of a solution, which is being titrated) for titration and determine equivalence point.  

Near equivalence point indicator should change color, so we must pick indicator who change color near pH of equivalence point.

8 0
3 years ago
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Compare natural selection to evolution. Be sure to justify your response in two or more complete sentences.
WINSTONCH [101]
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3 years ago
Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)
klio [65]

Answer:

Kp=3.07x10^6

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}

N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2

Consequently, the equilibrium constant is computed as:

Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6

Best regards.

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3 years ago
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