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Hunter-Best [27]
3 years ago
7

The volume of a gas at 17.5 psi decreases from 1.8L to 750mL. What is the new pressure of the gas in arm?

Chemistry
1 answer:
Maurinko [17]3 years ago
4 0

Answer:

P₂ = 2.88 atm

Explanation:

Given data:

Initial volume of gas = 1.8 L

Final volume = 750 mL

Initial pressure = 17.5 Psi

Final pressure = ?

Solution:

We will convert the units first:

Initial pressure = 17.5  /14.696 = 1.2 atm

Final volume = 750 mL ×1L/1000L = 0.75 L

The given problem will be solved through the Boly'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

1.2 atm × 1.8 L =  P₂  ×0.75 L

P₂ = 2.16 atm. L/ 0.75 L

P₂ = 2.88 atm

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3 years ago
Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th
ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}      .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol

The chemical equation for the reaction of ammonia and HCl follows:

NH_3+HCl\rightarrow NH_4^++Cl^-

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = \frac{1}{1}\times 0.0035=0.0035mol of ammonia

  • Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = \frac{1}{1}\times 0.0035=0.0035mol of ammonium ion

  • Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.8\times 10^{-5}

10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}

The chemical equation for the dissociation of ammonium ion follows:

NH_4^+\rightarrow NH_3+H^+

The expression of K_a for above equation follows:

K_a=\frac{[NH_3][H^+]}{[NH_4^+]}

We know that:

[NH_3]=[H^+]=x

[NH_4^+]=0.072M

Putting values in above expression, we get:

5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M

To calculate the pH concentration, we use the equation:

pH=-\log[H^+]

We are given:

[H^+]=6.32\times 10^{--6}M

pH=-\log (6.32\times 10^{-6})\\\\pH=5.20

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0
3 years ago
Which one of the samples contains the most molecules?
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Answer: D is right

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Despite of the substance

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A sample of a pure compound that weighs 59.8 g contains 27.6 g Sb (antimony) and 32.2 g F (fluorine). What is the percent compos
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Answer:

53.85%

Explanation:

Data obtained from the question include:

Mass of antimony (Sb) = 27.6g

Mass of Fluorine (F) = 32.2g

Mass of compound = 59.8g

Percentage composition of fluorine (F) =..?

The percentage composition of fluorine can be obtained as follow:

Percentage composition of fluorine = mass of fluorine/mass of compound x 100

Percentage composition of fluorine = 32.2/59.8 x 100

= 53.85%

Therefore, the percentage composition of fluorine in the compound is 53.85%

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3 years ago
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