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MAVERICK [17]
3 years ago
9

A total of 25.0 mL of 0.150 M potassium hydroxide (KOH) was required to neutralize 15.0 mL of sulfuric acid (H2SO4) of unknown c

oncentration. What is the concentration of the sulfuric acid? Balanced Equation: Acid-Base Titration Solution? Molar Ratio? Volume? Concentration? Acid (unknown) Base (known)
Chemistry
1 answer:
Kamila [148]3 years ago
6 0
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:

KOH + H₂SO₄ → H₂O + KHSO₄

If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:

0.025 L x 0.150 mol/L = .00375 mol KOH

0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄

We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:

0.00375 mol / 0.015 L = 0.25 mol/L

The concentration of H₂SO₄ being neutralized is 0.25 M.
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What is meant by conversion of energy
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If 0.500 mol neon at 1.00 atm and 273 K expands against a constant external pressure of 0.100 atm until the gas pressure reaches
trapecia [35]

Answer:

The work done on neon = -323 J

The internal energy change= -392.84 J

The heat absorbed by neon = -69.84 J

Explanation:

Step 1: Data given

Number of moles  = 0.500 moles

Pressure  = 1 atm

Temperature  = 273 Kelvin

The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.

a) calculate the work done on neon

W = -P(V2-V1)    

⇒ with P = the pressure = 0.1 atm

⇒ with V1 = the initial volume = nRTi /Pi

⇒ with V2 = the final volume = nRTf /Pf

W = -PnR((T2/P2) -(T1/P1))

⇒ with T2 = the final temperature = 210 K

 ⇒ with T1 = the initial temperature = 273 K

 ⇒ with P2 = the final pressure = 0.200 atm

 ⇒ with P1 = the initial pressure = 1.00 atm

W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))

W = -nR*(105 - 27.3)

W= -(0.500)*(8.314)*(77.7)

W = -323 J

b) calculate the internal energy change

E = (3/2)*nRT

ΔE = Ef - Ei

ΔE =(3/2)*nR(T2-T1)

⇒ with n= number of moles = 0.500 moles

⇒ with T2 =the final temperature = 210 K

⇒ with T1 = the initial temperature = 273 K

ΔE = (3/2)*(0.5)*(8.314)(210-273)

ΔE = -392.84 J

c) Calculate the heat absorbed by neon

ΔE = q + W

q = ΔE -W

⇒ with ΔE = -392.84 J

⇒ with W = -323 J

q = -392.84 J -( -323 J)

q =-392.84 J + 323 J

q = -69.84 J

4 0
3 years ago
The melting of a substance at its melting point is an isothermal process.
mamaluj [8]
The answer would be true
6 0
3 years ago
The change in entropy, Δ S ∘ rxn , is related to the the change in the number of moles of gas molecules, Δ n gas . Determine the
rewona [7]

Answer:

The entropy decreases.

Explanation:

The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where

Δngas = n(gaseous products) - n(gaseous reactants)

  • If Δngas > 0, the entropy increases
  • If Δngas < 0, the entropy decreases.
  • If Δngas = 0, there is little or no change in the entropy.

Let's consider the following reaction.

2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)

Δngas = 0 - 3 = -3, so the entropy decreases.

8 0
3 years ago
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