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Aleonysh [2.5K]
3 years ago
15

Why doesn't a convection cell form when the burner is turned off?​

Chemistry
1 answer:
kati45 [8]3 years ago
7 0

No convection cells are formed when the burner is off because there is no heat to make convection cells and you need heat in order to produce convection cells.

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Consider the rate law below.
Serga [27]

C. quadruples the rate

<h3>Further explanation</h3>

Given

The rate law :

R=k[A]²

Required

The rate

Solution

There are several factors that influence reaction kinetics :  

  • 1. Concentration  
  • 2. Surface area  
  • 3. Temperature  
  • 4. Catalyst  
  • 5. Pressure  
  • 6. Stirring  

The rate is proportional to the concentration.

If the concentration increased, the reaction rate will increase

The reaction is second-order overall(The exponent is 2)

The concentration of A is doubled, the reaction rate will increase :

r = k[A]² ⇒ r= k[2A]²⇒r=4k[A]²

<em>The reaction rate will quadruple.</em>

5 0
3 years ago
Some help I’ll give brainliest
daser333 [38]

Answer: Meteoroid

Explanation:

Dwarf planets are smaller than moons

Comets are smaller than dwarf planets.

Meteoroids, Meteors, and Meteorites are broken off pieces of comets.

Meteoroids are smaller than Asteroids

5 0
3 years ago
Classify..use the periodic table to classify the elements potassium, bromine a d argon according to how likely their atoms lose
iren2701 [21]

Answer:

Classification will be Potassium, Bromine, and Argon

Explanation:

  • Potassium is more likely to lose electrons and form positive ion
  • Bromine actually gain electrons and forms negative ion
  • Argon does not lose or gain electrons
6 0
3 years ago
"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

c_3=30.68J/g^oC

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0
4 years ago
The concept became more accepted by
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4 0
3 years ago
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