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____ [38]
3 years ago
11

Ethylene glycol (antifreeze) has a density of 1.11 g/cm3. what is the volume in liters of 3.46 kg of ethylene glycol?

Chemistry
1 answer:
Dovator [93]3 years ago
7 0

Density is a physical property which describes the mass of a substance per unit of volume of the substance. It is expressed as Density = m / V and it has units like g/cm^3. We use the density given to solve the problem.

3.46x10³ g / 1.11 g/cm³ = 3117.12 cm³  (1 L / 1000 cm³) =3.12 L

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Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the
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Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

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pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If you know pH and pka:

10^(pH-pka) = \frac{[A^-]}{[HA]}

The ratio will be: 10^(pH-pka)

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pH = pka + log₁₀ \frac{[A^-]}{[HA]}

0 = log₁₀ \frac{[A^-]}{[HA]}

10^0 = \frac{[A^-]}{[HA]}

1 = \frac{[A^-]}{[HA]}

As ratio is 1,  [conjugate base] = [acid] in solution.

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pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

  • At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

6 0
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Answer:

1.5 mol

Explanation:

Step 1: Given data

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Step 2: Calculate the moles corresponding to 33 L of argon at standard temperature and pressure (STP)

At STP, 1 mole of argon gas occupies 22.4 L.

33 L × 1 mol/22.4 L = 1.5 mol

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