Aballoon at 30.0°c has a volume of 222 ml. if the temperature is increased to 56.1°c and the pressure remains constant, what will the new volume be, in ml?

V₂ ≈ 241 mL

Math

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Chemistry

Unit 0

Gas Laws

Charles' Law:

Step 1: Define

V₁ = 222 mL

T₁ = 30.0 °C

T₂ = 56.1 °C

Step 2: Identify Conversions

Temp Conversion

Step 3: Convert

T₁ - 30.0 °C + 273.15 = 303.15 K

T₂ - 56.1 °C + 273.15 = 329.25 K

Step 4: Solve for V₂

Step 5: Check

We are given 3 sig figs. Follow sig fig rules and round.

241.113 mL ≈ 241 mL

one state would change   like louisiana to texas. thats a change

explanation:

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.