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Naily [24]
3 years ago
7

1. Describe the liquid state according to the kinetic-molecular theory.

Chemistry
1 answer:
Ahat [919]3 years ago
3 0

Answer:

Liquids have more kinetic energy than solids. When a substance increases in temperature, heat is being added, and its particles are gaining kinetic energy. Because of their close proximity to one another, liquid and solid particles experience intermolecular forces. These forces keep particles close together.

Explanation:

hope it helps U

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If you doubled the volume of a sample of gas and
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1125mL

Explanation:

this can be done using general gas law

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Order the following from least to most cellularly complex:
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organ, tissue, a cell, and a gorilla (i think)

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A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 20.00 gram sample of the alcohol p
bagirrra123 [75]

The empirical formula is C₂H₆O.

We must calculate the <em>masses of C, H, and O</em> from the masses given.

<em>Mass of C</em> =38.20 g CO₂ × (12.01 g C/44.01 g CO₂) = 10.424 g C

<em>Mass of H</em> = 23.48 g H₂O × (2.016 g H/18.02 g H₂O) = 2.6268 g H

<em>Mass of O</em> = Mass of compound - Mass of C - Mass of H

= (20.00 – 10.424 – 2.6268) g = 6.9487 g

Now, we must <em>convert these masses to moles</em> and <em>find their ratios</em>.

From here on, I like to summarize the calculations in a table.

<u>Element</u>  <u>Mass/g</u>    <u>Moles</u>    <u>Ratio</u>  <u>Integers</u>  

     C        10.424    0.8680  1.999        2

     H         2.6268  2.606    6.001        6

     O        6.9487  0.4343    1                1

The empirical formula is C₂H₆O.

7 0
3 years ago
A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system
AleksAgata [21]
To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

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Clearing the specific heat of the unknown substance:

c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}

c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}
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Round off the following number to four significant figures.<br> 273.15
Evgesh-ka [11]

Answer:

273.2

Explanation:

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