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lina2011 [118]
2 years ago
5

In the equilibrium system described by: PO43-(aq) + H2O(1) = HPO42-(aq) + OH-(aq) Brønsted-Lowry theory would designate: A) PO43

- and H20 as the bases B) H20 and OH as a conjugate pair C) HPO42- and OH"" as the acids D) HPO42- and H20 as a conjugate pair E) PO43-as amphiprotic
Chemistry
1 answer:
Kamila [148]2 years ago
6 0

Answer:

The correct option is:  B) H₂0 and OH⁻ as a conjugate pair

Explanation:

According to Brønsted-Lowry theory, the<u> </u><u>acids</u><u> are the chemical substances that form a conjugate base by donating a proton</u> and <u>bases</u><u> are the chemical substances that form conjugate acid by accepting a proton.</u>

In the given chemical reaction: PO₄³⁻(aq) + H₂O(l) ⇄ HPO₄²⁻(aq) + OH⁻(aq)

<u>According to Brønsted-Lowry theory, PO₄³⁻ and OH⁻ are bases. Whereas, H₂O and HPO₄²⁻ are acids.</u>

<u>Also, PO₄³⁻ and HPO₄²⁻ are the conjugate acid-base pair; and H₂O and OH⁻ are the conjugate acid-base pair.</u>

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A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0
2 years ago
Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0
ExtremeBDS [4]

Answer:

The freezing point will be 2.9^{0}C

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

Depressioninfreezing point=K_{f}Xmolality

Where

Kf= 20\frac{^{0}C}{m}

the molality is calculated as:

molality=\frac{moles_{solute}}{mass_{solvent}}

moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol

massofcyclohexane=25g=0.025Kg

molarity=\frac{0.0045}{0.025}=0.18m

Depression in freezing point = 20X0.18=3.6^{0}C

The new freezing point = 6.5^{0}C-3.6^{0}C=2.9^{0}C

5 0
2 years ago
3. Sharon woke up on a sunny morning and ate breakfast. Then she looked outside and saw tall, quickly-forming clouds. The clouds
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The change is that the air goes up then forms clouds.
7 0
2 years ago
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A lab director asks two students, Lydia and Damien, to each select a bottle of a concentrated weak base. When they reach the
Hunter-Best [27]

Answer:

First one is 5.0 M ammonia and the Second one ?

Explanation:

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2 years ago
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An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc
victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x10^{8} m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.

Answer: v% = 0.21 m/s

Explanation: To calculate the uncertainty, use <u>Heisenberg's Uncertainty Principle</u>, which states that:  ΔpΔx≥\frac{h}{4\pi }

where h is <u>Planck's constant</u> and it is equal to 6.626.10^{-34}m²kg/s.

Since p (momentum) is p = m.v:

mΔv.Δx ≥ \frac{h}{4\pi }

Δv = \frac{h}{4\pi.x.m }

Given that: r = x = 1.54.10^{-10}m and mass of an electron is m=9.1.10^{-31}kg

Δv = \frac{6.626.10^{-34} }{4.3.14.1.54.10^{-10}.9.1.10^{-31}}

Δv = 0.0376.10^{7}

As percentage of average speed:

Δv.\frac{1}{v}.100% = \frac{0.0376.10^{7} }{1.8.10^{8} }.10² = 0.021.10 = 0.21%

The least possible uncertainty in a speed of an electron is 0.21%.

5 0
2 years ago
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