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3 years ago

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3. Determine the Zeff value of Uranium, also indicate (based on your answer with mathematical calculations) which electron will

be closest to the nucleus of this element: 5s, 5p, 5f or 5d.

1 answer:

Ilya [14]3 years ago

3 0

Explanation:

Electronic configuration of uranium is given below -

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹⁴5d¹⁰6s²6p⁶5f³6d¹7s2²

Effective nuclear charge (Z eff) = Atomic number (Z) - Shielding constant (S)

<u> Value of Shielding constant (S) can be calculated by using slater's rule : </u>

S = 1 (0.35) + 9 (0.85) + 81 (1.00)

S = 0.35 + 7.65 + 81.00

S = 89

So,

Zeff = Z - S

Zeff = 92 - 89

<u> Zeff = 3 </u>

Hence, Effective nuclear charge (Zeff) of uranium = <u>3 </u>

5s electron is closest to the nucleus.

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Which reaction would cause an increase in entropy? A. SO2Cl2(g) SO2(g) + Cl2(g) B. PCl3(g) + Cl2(g) PCl5(g) C. 2CO(g) + O2(g) 2C

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

$2 NH_3$ ↔ $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

$[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

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The initial concentration of nitrogen gas is mathematically represented a

$[N_2] = \frac{n_2}{V_2}$

$[N_2] = 0.173 \ M$

So looking at the equation

Initially (Before reaction)

$NH_3 = 0.387 \ M$

$N_2 = 0 \ M$

$H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

$NH_3 = -2 x$ (this implies that it losses two moles of concentration )

$N_2 = + x$ (this implies that it gains 1 moles)

$H_2 = +3 x$ (this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

$NH_3 = 0.387 -2x$

$N_2 = x$

$H_2 = 3 x$

Now since

$[NH_3] = 0.387 \ M$

$x= 0.387 \ M$

$H_2 = 3 * 0.173$

$H_2 = 0.519 \ M$

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

$K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

$K_c = 14.39$

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A chemist adds 0.60L of a 0.20/molL sodium thiosulfate Na2S2O3 solution to a reaction flask. Calculate the millimoles of sodium

Serggg [28]

Answer:

1.2×10² mmole of Na₂S₂O₃

Explanation:

From the question given above, the following data were obtained:

Volume = 0.6 L

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Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity = mole /Volume

0.2 = Mole of Na₂S₂O₃ / 0.6

Cross multiply

Mole of Na₂S₂O₃ = 0.2 × 0.6

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1 mole = 1000 mmol

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0.12 mole = 0.12 mole × 1000 mmol / 1 mole

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Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃

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