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

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.