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3 years ago

What is the hydronium h3o+ concentration of a solution with a pH pf 3

Chemistry

1 answer:

arlik [135]3 years ago

7 0

Answer:

1x10^-3M

Explanation:

Data obtained from the question include:

pH = 3

[H3O+] =?

The concentration of the hydronium ion [H3O+] can be obtained by using the formula pH = - Log [H3O+] as illustrated below:

pH = - Log [H3O+]

3 = - Log [H3O+]

Divide through by - 1

- 3 = Log [H3O+]

Take the anti-log of - 3

[H3O+] = 1x10^-3M

Therefore, the concentration of the hydronium ion [H3O+] is 1x10^-3M

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1 year ago

Copper metal has a specific heat of 0.385 J/goC. Calculate the amount of heat required to raise the temperature of 22.8 g of Cu

Liula [17]

Answer is equal to 7.51 kJ

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2 years ago

The number of atoms or molecules in a mole is known as the __________ constant.

AnnZ [28]

Answer: Avogrado's Constant

Explanation:

One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles.

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2 years ago

If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?

aksik [14]

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

Mn(OH)₂ → Mn⁺² + 2OH⁻

[Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

Ksp = [Mn⁺²][OH-]²

[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

[OH⁻]² = 10⁻³

[OH⁻] = (10⁻³)¹⁽²

[OH⁻] = 0.0316 M

Calculate the pOH

pOH = -log [OH⁻]

pOH = -log [0.0316]

pOH = 1.5

Now calculate pH

pH = 14 - pOH

pH = 14 - 1.5

pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

#SPJ4

7 0

1 year ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

5 0

3 years ago