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3 years ago

Consider the following reaction at 298K.

Chemistry

1 answer:

dmitriy555 [2]3 years ago

8 0

Answer:

Eªcell > 0; n = 2

Explanation:

The reaction:

I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)

Is product favored.

A reaction that is product favored has ΔG < 0 (Spontaneous)

K > 1 (Because concentration of products is >>>> concentration reactants).

Eªcell > 0 Because reaction is spontaneous.

And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:

<h3>Eªcell > 0; n = 2</h3>

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A sample of 0.281 gg of an unknown monoprotic acid was dissolved in 25.0 mLmL of water and titrated with 0.0950 M NaOH NaOH. The

tensa zangetsu [6.8K]

Answer:

98.6 g/mol.

Explanation:

Equation of the reaction

HX + NaOH--> NaX + H2O

Number of moles = molar concentration × volume

= 0.095 × 0.03

= 0.00285 moles

By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.

Molar mass = mass ÷ number of moles

= 0.281 ÷ 0.00285

= 98.6 g/mol.

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4 years ago

At its closest approach to the sun, Pluto is less than 30 AU from the sun. Because Pluto is approaching that point now, which pl

MA_775_DIABLO [31]

Neptune: its the 8th planet and since Pluto isn't a planet anymore Neptune is the farthest planet from the sun with a distance of around 2.8 billion miles (4.5 billion km) or (30.07 AU)

6 0

3 years ago

If the reactants are 10 kg what is the mass of the products

nlexa [21]

If the mass of both the reactants is 10kg then the mass of the products also equals 10kg.

It is due to the law of conservation of mass.

Mass can neither be created nor be destroyed.

7 0

3 years ago

A BaSO4 slurry is ingested before the gastrointestinal tract is x-rayed because it is opaque to x-rays and defines the contours

NNADVOKAT [17]

Explanation:

The given reaction equation is as follows.

$BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO_{4}^{2-}(aq)$

The value of $\Delta G^{o}$ = 59.1 kJ/mol

We know that ,

$\Delta G^{o} = -RT ln K_{sp}$

or, $ln K_{sp} = -(\frac{\Delta G^{o}}{RT})
$

= -(\frac{59.1 kJ/mol}{(8.314 \times 10^{-3} kJ/mol.K \times 310 K))}[/tex]

= -22.93

or, $K_{sp} = e^{-22.93}$

= $1.1 \times 10^{-10}$

$K_{sp} = [Ba^{2+}][ SO_{4}^{2-}]$

Therefore, $[Ba^{2+}] =\sqrt{K_{sp}}$

= $\sqrt{ 1.1 \times 10^{-10}}$

= $1.05 \times 10^{-5} M$

Therefore, we can conclude that the value of $[Ba^{2+}]$ in the intestinal tract is $1.05 \times 10^{-5} M$ .

7 0

3 years ago

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Ksivusya [100]

The answer is C.) Pair it with a hydrogen half-cell

6 0

3 years ago