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astraxan [27]
3 years ago
11

If a certain gas occupies a volume of 19 L when the applied pressure is 9.5 atm, find the pressure when the gas occupies a volum

e of 4.8 L
Chemistry
1 answer:
kakasveta [241]3 years ago
3 0

Answer: 37.6 atm

Explanation:

Given that,

Initial volume of gas (V1) = 19L

Initial pressure of gas (P1) = 9.5 atm

Final volume of gas (V2) = 4.8L

Final pressure of gas (P2) = ?

Since pressure and volume are given while temperature remains the same, apply the formula for Boyle's law

P1V1 = P2V2

9.5 atm x 19L = P2 x 4.8L

180.5 atm•L = 4.8L•P2

Divide both sides by 4.8L

180.5 atm•L/4.8L = 4.8L•P2/4.8L

37.6 atm = P2

Thus, the final pressure is 37.6 atmospheres.

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The percent yield of this reaction is consistently 92.0%. CH4(g) + 4 S(g) → CS2(g) + 2 H2S(g) How many grams of sulfur would be
dybincka [34]

Answer:

132.17 g

Explanation:

The reaction given , in the question is -

    CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S  ( g )

From the reaction , 4 mole of S is required for the production of 1 mole of  CS₂ .

since ,

Moles of  CS₂  = given mass of CS₂ / Molecular weight  of  CS₂

Since ,

the Molecular weight of  CS₂ = 76

Given ,  mass of CS₂ =  72.57 g

Moles of  CS₂ = 72.57  / 76 = 0.95 mol

Since ,

The yield is 92.0 % .

Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles

Mass of S required = 4.13 * 32 = 132.17 g .

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What makes the results of a scientific experiment accurate?
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Explanation:

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Where does radioactivity come from?
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Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)
miss Akunina [59]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

  • Adding solute to water elevates the boiling point.
  • The elevation in boiling point (ΔTb) can be calculated using the relation:

<em>ΔTb = i.Kb.m,</em>

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

  • van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

<u><em>(1) 1.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

<u><em>(2) 2.0 M KCl(aq):</em></u>

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

<u><em>(3) 1.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

<u><em>(4) 2.0 M CaCl₂(aq):</em></u>

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

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<em></em>

6 0
3 years ago
calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate
kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

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