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alexandr1967 [171]
3 years ago
8

Which best describes the transfer of heat energy through conduction? A. Heat energy is transferred from molecule to molecule by

collision. B. Cool, high pressure air pushes warm, low pressure air out of the way. C. Energy absorbed from the sun is converted into heat energy. D. Heat energy is trapped by the atmosphere.
Chemistry
1 answer:
TiliK225 [7]3 years ago
5 0
It should be option (A) because conduction is the transfer of heat energy from one place to another by vibrations/motion of the molecules. I hoped I helped.
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A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
3 years ago
What is the conclusive evidence that mixing sugar in the water is a physical change
djyliett [7]
THE DEFINITION OF PHYSICAL CHANGE: Physical changes are changes affecting the form of a chemical substance, but not its chemical composition. Physical changes are used to separate mixtures into their component compounds, but can not usually be used to separate compounds into chemical elements or simpler compounds. so the answer is that the form of the sugar is changing in water but if you boiled the water till its all evaporated all that will be left is the sugar
8 0
4 years ago
How are models useful in describing how molecules are formed
Nady [450]

Answer:

The purpose of molecular modeling is to provide a three-dimensional image (either physical or software-based) that allows a chemist to better see the manner in which atoms and molecules can interact. These models can be used to interpret existing observations or to predict new chemical behavior.

Explanation:

It can describe shape and how they connect while forming the electrons.

6 0
3 years ago
Which of the following methods can be used to identify the suspected use of heat lamps in a house without entering the house?
LuckyWell [14K]

Answer:

b I see some of ask you to do it

7 0
3 years ago
What is the freezing point (in degrees Celcius) of 4.14 kg of water if it contains 235.1 g of butanol, C 4 H 9 O H
Karolina [17]

Answer:

Explanation:

Molal freezing point depression constant of butanol Kf = 8.37⁰C /m

ΔTf = Kf x m , m is no of moles of solute per kg of solvent .

mol weight of butanol = 70 g

235.1 g of butanol = 235.1 / 70 = 3.3585 moles

3.3585 moles of butanol dissolved in 4.14 kg of water .

ΔTf = 8.37 x 3.3585 / 4.14

= 6.79⁰C

Depression in freezing point = 6.79

freezing point of solution = - 6.79⁰C .

5 0
3 years ago
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