Chemistry, 27.05.2021 06:00 JN02

The pressure of a certain amount of gas in a 1.25 L balloon is 760 torr. What would the pressure of that balloon be if it's volume was increased to 2.15 L?

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Answer from: rleiphart1

440 torr

General Formulas and Concepts:

Chemistry

Ideal Gas Law

Boyle's Law: P₁V₁ = P₂V₂

P₁ is pressureV₁ is volumeP₂ is new pressureV₂ is new volume

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Equality Properties

Multiplication Property of Equality Division Property of Equality Addition Property of Equality Subtraction Property of Equality Explanation:

Step 1: Define

Identify

[Given] V₁ = 1.25 L

[Given] P₁ = 760 torr

[Given] V₂ = 2.15 L

[Solve] P₂

Step 2: Solve for P₂

Substitute in variables [Boyle's Law]: (760 torr)(1.25 L) = P₂(2.15 L)Multiply: 950 torr · L = P₂(2.15 L)[Division Property of Equality] Divide 2.15 L on both sides: 441.86 torr = P₂Rewrite: P₂ = 441.86 torr

Step 3: Check

Follow sig fig rules and round. We are given 2 sig figs as our lowest.

441.86 torr ≈ 440 torr

Answer from: Quest

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.

chemical decomposition is opposite of chemical synthesis.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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