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2 years ago

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How many moles of boron (B, 10.81 g/mol) are in 37.8 grams of boron? [?] moles

1 answer:

madreJ [45]2 years ago

3 0

Answer:

37.8g/ 10.81g/mol = 3.4968...moles

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The effective nuclear charge is the amount of charge experienced by an electron taking into account any shielding effects from o

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Answer:

The nuclear charge increases from boron to carbon, but there is no additional shielding( that is no additional shells).

Explanation:

First of all, we must know the electron configuration of carbon and boron.

Boron- 1s2 2s2 2p1

Carbon- 1s2 2s2 2p2

Moving from boron to carbon, the effective nuclear charge increases without a corresponding increase in the number of shells. Remember that shielding increases with increase in the number of intervening shells between the outermost electron and the nucleus. Since there isn't an increase in shells, boron experience a lower screening effect.

From

Zeff= Z- S

The Z for carbon is 6 while for boron is 5 even though both have the same number of screening electron S(4 screening electrons). Hence it is expected the Zeff(effective nuclear charge) for boron will be less than that of carbon.

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A scientist observes how a particular South American bird cares for its young. Why is a field study an appropriate type of scien

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Who first argued that all matter is made up of particles so small that they could not possibly be divided

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3 years ago

How long is 5 half lives if the half life is 2.6 hours? Remember to round to the correct number of significant figures and use u

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3 years ago

In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher

trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of $BC_2$ that should used originally is $C_Z_o = 0.4492M$

Explanation:

From the question we are told that

The necessary elementary step is

$2BC_2 ----->4C + B_2$

The time taken for sixth of 0.5 M of reactant to react $t = 9 hr$

The time available is $t_a = 3.5 hr$

The desired concentration to remain $C = 0.42M$

Let Z be the reactant , Y be the first product and X the second product

Generally the elementary rate law is mathematically as

$-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}$

Where k is the rate constant , $C_Z$ is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

For second order reaction

$\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt$

Where $C_Z_o$ is the initial concentration of Z which a value of $C_Z_o = 0.5M$

From the question we are told that it take 9 hours for the concentration of the reactant to become

$C_Z = C_Z_o - \frac{1}{6} C_Z_o$

$C_Z = 0.5 - \frac{0.5}{6}$

$= 0.4167 M$

So

$\frac{1}{0.4167} - \frac{1}{0.50} = 9 k$

$0.400 = 9 k$

=> $k = 0.044\ L/ mol \cdot hr^{-1}$

For $C_Z = 0.42M$

$\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044$

$2.38 - 0.154 = \frac{1}{C_Z_o}$

$2.226 = \frac{1}{C_Z_o}$

$C_Z_o = \frac{1}{2.226}$

$C_Z_o = 0.4492M$

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3 years ago