All categories

2 years ago

What is the name of this compound?

Chemistry

1 answer:

Monica [59]2 years ago

7 0

Answer:

THERE IS NOTHING MENTION HERE HOW CAN ANYONE KNOW ABOUT IT?]

Explanation:

You might be interested in

Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole

kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ $\frac{[A^-]}{[HA]}$

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where CH₃COOH is the acid and CH₃COO⁻ is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ $\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}$

a) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[2 mol]}$

pH = 4,75

b) The pH is:

pH = 4,75 + log₁₀ $\frac{[2 mol]}{[1mol]}$

pH = 5,05

I hope it helps!

7 0

3 years ago

Will reward brainliest !!!

Anna [14]

The best answer choice here would be 'Combination'

8 0

3 years ago

Read 2 more answers

If an acid/base dissociates completely in water what type of substance is it? Does this mean the acid/base is dangerous?

hram777 [196]

Answer:

1 strong acid

2 yes they are dangerous

Explanation:

Since nearly all of it is dissociated in water, it is called a strong acid.

2 yes Concentrated strong acids can cause severe and painful burns. The pain is due in part to the formation of a protein layer, which resists further penetration of the acid

3 0

3 years ago

1. What is the oxidation number for the silver ion in tarnish?

spayn [35]

Tarnish is Ag2S-silver sulfide and the oxidation state of silver is +1

7 0

2 years ago

Calculate the values of ΔU, ΔH, and ΔS for the following process:

ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

H2O(l) → H2O(l) → H2O(steam)

298.15K, 1atm ΔHp 373.15K,1atm ΔHv 373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 0

3 years ago