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Lemur [1.5K]
3 years ago
14

Match the definitions below with the correct component of the scientific method:

Chemistry
1 answer:
laiz [17]3 years ago
8 0
A and 3

B and 6

C and 5

D and 1

E and 2

F and 4
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The difference between the density of gas and the density of liquid
eduard

The molecules of a liquid substance are closely packed together to each other. So as a result, liquids are denser than gases.

<h3>What is the difference between the density of liquid and gas?</h3>

A mass of gas will have a much larger volume compared to the same mass of liquid. This is because it has a much lower density. The density of gaseous oxygen is 0.0014 g/cm3. Density is ρ=Mass Volume. We know that gas will uniformly occupy more space than liquid whatever volume is available to it. On the other hand, solids and liquids, are closely packed as compared to gas and are high-density materials where ρ is relatively constant.

So we can conclude that the molecules of a liquid substance are closely packed together with each other. So as a result, liquids are denser than gases.

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2 years ago
Dan got into his car in the middle of summer. When he sat on the hot black leather seats, he burned his legs. His sister said th
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The correct answer is D) Dan's sister was correct because Dan's legs touched the car seats. That is an indicator of heat transfer by conduction.

Conduction only happens when a heated object touches a non-heated (or not as heated) object. Radiation did cause the car to become hot, but conduction caused Dan to get burned.

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31. Adding an additional oxygen atom to 02 creates....
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A chemist reported that 100 gallons of a gas were available in a laboratory. Which property of the gas did the chemist report?
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It would be volume

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What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S
Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat,  the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

  • <u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • c= Heat Capacity of Liquid= 4.184 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 \frac{J}{gC}× 185.5 g× (- 25.6 °C)

Solving:

<u><em>Q1= -19,868.98 J</em></u>

  • <u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× \frac{1mol}{18 grams}= 10.30 moles, where 18 \frac{g}{mol} is the molar mass of water, that is, the amount of mass that a substance contains in one mole.

ΔHfus= 6.01 \frac{kJ}{mol}

Replacing:

Q2= 10.30 moles×6.01 \frac{kJ}{mol}

Solving:

<u><em>Q2=61.903 kJ= 61,903 J</em></u>

  • <u><em>0 °C to -10.70 °C</em></u>

Similar to sensible heat previously calculated, you know:

  • c = Heat Capacity of Solid = 2.092 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C

Replacing:

Q3= 2.092 \frac{J}{gC} × 185.5 g× (-10.70) °C

Solving:

<u><em>Q3= -4,152.3062 J</em></u>

<h3>Total heat required</h3>

The total heat required is calculated as:  

Total heat required= Q1 + Q2 +Q3

Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J

<u><em>Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ</em></u>

In summary, the amount of heat required is 37.88 kJ.

Learn more about calorimetry:

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2 years ago
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