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zhenek [66]
3 years ago
10

A balloon has a volume of 3.50 liters at 20.0°C. What will be the volume of the balloon at 100.0°C? Assume the pressure doesn't

change and the amount of gas in the balloon doesn't change O a. 0.224 Ob.2.75L O c. 17.5L O d. 65.45 Oe. None of the above
Chemistry
1 answer:
Nikitich [7]3 years ago
5 0

<u>Answer:</u> The volume of balloon at 100°C is 4.46 L

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=3.50L\\T_1=20^oC=(20+273)K=293K\\V_2=?L\\T_2=100^oC=(100+273)K=373K

Putting values in above equation, we get:

\frac{3.5L}{293K}=\frac{V_2}{373K}\\\\V_2=4.46L

Hence, the volume of balloon at 100°C is 4.46 L

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A 59.1g sample of aluminum is put into a calorimeter (see sketch at right) that contains 250.0g of water. The aluminum sample st
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Answer:

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

Explanation:

Step 1: Data given

Mass of aluminium = 59.1 grams

Mass of water = 250.0 grams

Initial temperature of aluminium = 91.3 °C

Initial temperature of water = 16.0 °C

Final temperature = 19.5 °C

Pressure remains constant

Specific heat capacity of water = 4.186 J/g°C

Step 2: Calculate specific heat of aluminium

Heat lost = heat gained

Qlost = -Q heat

Q = m*c*ΔT

heat aluminium = - heat water

m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)

⇒m(aluminium) = mass of aluminium = 59.1 grams

⇒c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 91.3 = -71.8 °C

⇒ m(water) = 250.0 grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 16.0 = 3.5 °C

59.1 * c(aluminium) * -71.8 °C = 250.0 * 4.186 J/g°C * 3.5 °C

c(aluminium) = 0.863 J/g°C

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

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3 years ago
Define Nuclear Fission with an example of it​
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Answer:

Fission is the splitting of an atomic nucleus into two or more lighter nuclei accompanied by energy release. ... The energy released by nuclear fission is considerable. For example, the fission of one kilogram of uranium releases as much energy as burning around four billion kilograms of coal

Explanation:

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Explanation:

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Which of these statements are true? Select all that apply.
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The AH+ for Br2() is 0 kJ/mol.

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Explanation:

7 0
3 years ago
29. If a walnut is burned in a
harkovskaia [24]

Energy were released from the walnut, q = 1,673.6 J

<h3>Equation :</h3>

To find the energy using formula,

q = mcΔt

where,

q is charge

m is mass

c is specific heat of water

Δt is change in temperature

So, given

t₁ = 50°C

t₂ = 60°C

m = 40g

c = 4.184 J/g

Now putting the values known,

We get,

q = mc(t₂ - t₁)

q = 40g x 4.184 J/g x  (60 - 50)

q = 167.36 J x 10

q = 1,673.6 J

<h3>What is heat energy?</h3>

Heat is the thermal energy that is transferred when two systems with different surface temperatures come into contact. Heat is denoted by the letters q or Q and is measured in Joules.

To know more about specific heat :

brainly.com/question/11297584

#SPJ9

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