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just olya [345]
3 years ago
15

For ethanol, propanol, and n-butanol the boiling points, surface tensions, and viscosities all increase. what is the reason for

this increase?
Chemistry
1 answer:
Contact [7]3 years ago
3 0
Moving from Ethanol through Propanol to Butanol the physical properties like boiling points, surface tension and viscosity increases because of the increases in intermolecular interactions between the molecules of given compounds.

Explanation:
                   Ethanol, propanol and butanol all have hydroxyl groups in common, means all have hydrogen bond intractions between their molecules. So, taking the hydrogen bonding interaction constant we are left with only the difference in the number of carbon atoms.
                    Butanol has the greatest physical properties than other two because it has four carbon atom chain. So, as we know the London Dispersion forces or Van der Waal forces increases with increase in molecular size and chain length of hydrocarbon.
                    Therefore, the strength of London forces is greater in butanol than other two while ethanol has the smallest chain comparatively hence, lowest physical properties.  
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3 years ago
Identify the term that applies to each definition.
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Answer:

A. Reference blank

B. Cuvettes

C. Transmittance

D. Absorbance

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Explanation:

A reference blank is a sample prepared using the solvent and any other chemicals in the sample solutions, but not the absorbing substance.

A square-shaped container, typically made of quartz, designed to hold samples in a spectrophotometer is known as Cuvettes.

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Antibiotics do not reduce the number of helpful and harmful bacteria in the microbiome.
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Explanation:

7 0
3 years ago
Read 2 more answers
1.Give an example of a compound machine (1pt). Explain how at least two simple machines are part of this compound machine (2pts)
Law Incorporation [45]
1) An example of a compound machine could be a pair of Scissors. Their are two different simple machines in the Scissors which make up the compound machine. Both of them being a Lever, and a Fulcrum.

Hope this helps!
4 0
3 years ago
Can someone help?? This is really hard.
Sergeu [11.5K]

Answer:

See Explanation

Explanation:

Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.

2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4

2^1 = 2^1

The rate of reaction is first order with respect to Hbn

Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.

1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4

3^1 = 3^1

The reaction is also first order with respect to CO

b) The overall order of reaction is 1 + 1=2

c) The rate equation is;

Rate = k [CO] [Hbn]

d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4  /[5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4/6.7 * 10^-7

k = 4.7 * 10^2 mmol-1 L s-1

e) The reaction occurs in one step because;

1) The rate law agrees with the experimental data.

2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.

8 0
3 years ago
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