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2 years ago

Calculate the new pressure if a 2300 mL of a gas at a pressure of 1.88 atm is allowed to contract to a volume of 2.21 L.

Chemistry

1 answer:

olya-2409 [2.1K]2 years ago

6 0

Answer:

V₂ = 0.958 atm

Explanation:

Given data:

Initial volume = 2300 mL (2300/1000 = 2.3 L)

Initial pressure = 1.88 atm

Final volume = (2.21 L + 2.3L) = 4.51 L

Final pressure = ?

Solution;

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

1.88 atm × 2.3 L = P₂ × 4.51 L

V₂ = 1.88 atm × 2.3 L / 4.51 L

V₂ = 4.32 atm.L / 4.51 L

V₂ = 0.958 atm

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A balloon contains 0.140 molmol of gas and has a volume of 2.78 LL . If an additional 0.152 molmol of gas is added to the balloo

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The final volume will be 5.80 L

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Step 1: Data given

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Step 2: Calculate the final volume

V1/n1 = V2/n2

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⇒ with V2 = The new volume = TO BE DETERMINED

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II. Ionic Equations

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Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .

Simplify the coefficients:

$\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

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