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MatroZZZ [7]
3 years ago
8

Why do convection currents form

Chemistry
1 answer:
Scrat [10]3 years ago
3 0
Convection currents can happen in liquids and gases. For example when air is heated it becomes less dense and rises. This means that cooler more dense air comes down and is also heated rising up above more cooler air etc. This creates a convection current.
Here is a diagram of a hot water tank heater:
(sorry it isn't that great I did it myself just now)

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Octane (C8H18) is a component of gasoline that burns
MatroZZZ [7]

Answer:

hold on this will take me awhile

Explanation:

4 0
3 years ago
What would be the greatest difference in using an open ceramic coffee mug rather than an insulated mug with a lid as a calorimet
adoni [48]
The answer is number 4 or the exchange of energy with the surroundings. Calorimetry is a measurement of energy that is formed or absorbed in a certain process. The calorimeter is the instrument used in order to measure the energy. It is recommended that a calorimeter should be a closed system so as to measure precisely the energy and avoid or lessen the exchange of energy with the surroundings. Thus, comparing an open ceramic mug and an insulated mug with a lid, the greatest difference is the energy lost to the surroundings. 
4 0
3 years ago
Read 2 more answers
a chemical reaction produced 2.50 moles of nitrogen gas. what volume in liters, does this gas sample occupy at STP? (show your w
Harrizon [31]

The volume of N₂ at STP=56 L

<h3>Further explanation</h3>

Given

2.5 moles of N₂

Required

The volume of the gas

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, the volume per mole of gas or the molar volume-Vm is 22.4 liters/mol.

So for 2.5 moles gas :

\tt 2.5\times 22.4=56~L

5 0
3 years ago
g A gas is compressed in a cylinder from a volume of 20.0 L to 2.0 L by a constant pressure of 10.0 atm. Calculate the amount of
Lemur [1.5K]

Answer:

The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.

Explanation:

Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).

The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:

W system= -p*∆V

Where:

  • W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)
  • p: Pressure. Its unit of measurement in the International System is the pascal (Pa)
  • ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)

In this case:

  • p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)
  • ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)

Replacing:

W system= -1.013*10⁶ Pa* (-0.018 m³)

Solving:

W system= 18234 J

<u><em>The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.</em></u>

5 0
3 years ago
Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

4 0
3 years ago
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