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qaws [65]
3 years ago
5

Assuming a 100% dissociation, what is the freezing point and boiling point of 2.59 m K3PO4(aq).

Chemistry
2 answers:
liubo4ka [24]3 years ago
3 0
Assuming 100% dissociation of K3PO4 you get each molecule of K3 PO4 dissociates into four particles: 3 K4 cations and 1 PO4 - anion. That means that Van't Hoff factor, i, is 4. i = 4. So, the decrease in freezing point of the water solution is Delta Tf = i * Kf * m. Where Kf is the molal cryoscopic constant of water = 1.86 °C /m; and the molality m = 2.59 m => Delta Tf = 4 * 1.86 °C / m * 2.59 m = 19.3 °C. And the freezing point is the normal freezing points less 19.3°C = 0°C - 19.3°C = - 19.3°C. The increase in the boiling point Delta Tb = i * kb * m = 4 * 0.512 °C /m * 2.59 m = 5.3°C => <span>Tb = normal boiling point + 5.3°C = 100°C + 5.3°C = 105.3°C.</span>
Romashka [77]3 years ago
3 0

Answer: Tb = i * kb * m = 4 * 0.512 °C /m * 2.59 m = 5.3°C => Tb = normal boiling point + 5.3°C = 100°C + 5.3°C = 105.3°C.

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monitta
N=6.98*10²⁴
Nₐ=6.022*10²³ mol⁻¹

n(Mg)=N/Nₐ

m(Mg)=n(Mg)M(Mg)=M(Mg)N/Nₐ

m(Mg)=24.3g/mol*6.98*10²⁴/(6.022*10²³mol⁻¹)=281.7 g
5 0
3 years ago
Match: Substance ....... Definition Lewis acid A. Provides H+ in water Bronsted-Lowry base B. Provides OH - in water Arrhenius b
Igoryamba

Answer: F.    Electron pair acceptor

Explanation:

A Lewis acid can be properly defined as any substance such as  H+ (hygrogen ion) that can accept a pair of electron.

While a Lewis base is any substance such as (OH-) that can donate a pair of electron.

In the neutralization reaction between an acid ( H+ ) and a base (OH-). Hydrogen ion (H+ ) is the Lewis acid because it accepts an electron pair from (OH-).

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3 0
3 years ago
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
inn [45]

Answer:

3.47 ×10^-10

Explanation:

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A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

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log K = -9.4595

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K= 3.47 ×10^-10

4 0
3 years ago
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Answer:

54.99% yield

Explanation:

percent yield is just the amount you obtained over the amount expected times 100%.

(experimental value/theoretical value) x 100%

= (107.9 g/196.2 g) x 100%

=54.99% yield

5 0
3 years ago
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Answer:

Its B. Dilute acid + carbonate

Explanation:

hope it helps !!

3 0
3 years ago
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