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pogonyaev
3 years ago
13

Which of the following is likely to have the lowest viscosity? hot oil room temperature oil room temperature water below room te

mperature oil
Chemistry
2 answers:
goblinko [34]3 years ago
5 0

Answer: Option (a) is the correct answer.

Explanation:

Viscosity is defined as the ability of a liquid to resist its flow.

When there is more force of attraction between the molecules of a liquid then the liquid will be more viscous.

Out of the given options, hot oil will have be less viscous as due to the heat its molecules will be far apart from each other. Therefore, there will be less force of attraction between its molecules. So, it will flow easily.

Hence, we can conclude that out of the given options hot oil is likely to have the lowest viscosity.

zzz [600]3 years ago
4 0

Viscosity, is a state of matter where it resists flow or may be in thick semifluid form because of friction acting inside the matter.In this case, hot oil room temperature would lessen the viscosity of the oil (animal oil). Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
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Primary active transport                                                                                              

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52.1 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 38.5 mL of 0.415 M NaCl. The equation for the precipitate reaction is: Pb(NO3)
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Answer:

0.0585 M

Explanation:

  • Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)

First we <u>calculate the inital number of moles of each reagent</u>, using the <em>given volumes and concentrations</em>:

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  • 0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl

Then we <u>calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl</u>, using the <em>stoichiometric coefficients of the reaction</em>:

  • 16.0 mmol NaCl * \frac{1mmolPb(NO_3)_2}{2mmolNaCl} = 8.00 mmol Pb(NO₃)₂

Now we <u>calculate the remaining number of Pb(NO₃)₂ moles after the reaction</u>:

  • 13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂

Finally we <em>divide the number of moles by the final volume</em> to <u>calculate the concentration</u>:

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