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lawyer [7]
3 years ago
13

How could two objects have the same temperature but different thermal energies? One object could have more particles and greater

total kinetic energy. One object could have more particles and lesser total kinetic energy. One object could have more calories. One object could have more heat.
Chemistry
1 answer:
Tju [1.3M]3 years ago
6 0

Answer:

One object could have more particles and greater total kinetic energy

Explanation:

The higher the temperature the more the particles. So, when we have high temperature, there is more particles interacting.

Temperature can simply be defined or gotten by taking the average of of the kinetic energy of the particles in the object that is the keywords here are TAKING THE AVERAGE KINETIC ENERGY

Thermal energy is the energy that can be gotten from adding up all the total kinetic energy of the particles in the object. So, the keywords here are; ADDING UP THE TOTAL KINETIC ENERGIES.

So, when two particles of the same temperature have the different thermal energy it means that One object could have more particles and thus having greater total kinetic energy.

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3 0
3 years ago
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5000kg of ammonium nitrate per square kilometer of cornfield per year. how much nitric acid would be needed to make the fertiliz
Serggg [28]

NH_3+HNO_3-> NH_4NO_3

1 mole of nitric acid produce 1 mole of ammonium nitrate.

moles in 5000 kg of ammonium nitrate :

n=\dfrac{5000\times 1000}{80}\\\\n=62500\ moles ( molecular mass of ammonium nitrate is 80 gm/mol )

So, number of moles of nitric acid required are also 62500 moles.

Mass of 62500 moles of nitric acid :

mass = 62500\times 63 \\\\mass = 3937500\ gram\\\\mass = 393.75\ kg

Hence, this is the required solution.

5 0
3 years ago
Write a word equation for the combustion of propane???<br> HELP????
frutty [35]

Answer:

1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.

Explanation:

The word equation for the combustion of propane can be obtained from the chemical equation;

      C₃H₈   +    5O₂    →   3CO₂   +   4H₂O  

The word equation is therefore:

 1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.

For such a combustion reaction, carbon dioxide and water are produced in the process.

8 0
3 years ago
What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?
Vsevolod [243]

The dilution formula can be used to find the volume needed

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

c1 - 0.33 M

c2 - 0.025 M

v2 - 25 mL

Substituting these values in the equation

0.33 M x v1 = 0.025 M x 25 mL

v1 = 1.89 mL

Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution

7 0
3 years ago
A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°C. Calculate the
Kryger [21]

Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

The specific heat capacity of the metal is 0.268 J/g°C

5 0
3 years ago
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