Balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room? Use

The answer to your question is V2 = 3.7 L

Explanation:

Data

Volume 1 = 3.5 L

Temperature 1 = 25°C

Temperature 2 = 40°C

Pressure = cte.

Volume 2 = ?

Formula

To solve this problem use the Charles' law

             V1/T1 = V2/T2

- Solve for V2

             V2 = V1T2 / T1

- Convert temperature to °K

Temperature 1 = 25 + 273 = 298°K

Temperature 2 = 40 + 273 = 313°K

- Substitution

              V2 = (3.5)(313) / 298

- Simplification

              V2 = 1095.5 / 298

- Result

              V2 = 3.7 L

The new volume of the balloon is 3.68 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 3.50 L

The temperature = 25.0 °C = 298 K

The temperature changes to 40.0 °C = 313 K

The pressure remains constant at 1.0 atm

Step 2: Calculate the new volume of the balloon

V1/T1 = V2/T2

⇒with V1 = the initial volume of the balloon = 3.50 L

⇒with T1 = the initial temperature = 298 K

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the new temperature = 313 K

3.50 L / 298 K = V2 / 313 K

V2 = 3.68 L

The new volume of the balloon is 3.68 L

the law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. because atoms are only rearranged in a chemical reaction, there must be the same number of sodium atoms and chlorine atoms in both the reactants and products

well in order to answer this question you have to provide more information < 3