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9966 [12]
2 years ago
9

Using principles of atomic structure, explain why the atomic radius of Te is less than the ionic radius of Te2− . Photoelectron

spectroscopy data for the 1s sublevel of Te and the 1s sublevel of O are represented below. In terms of Coulomb's law and atomic structure, explain why the peak for O is positioned so far to the right of the peak for Te .
Chemistry
1 answer:
ValentinkaMS [17]2 years ago
3 0

Answer:

See explanation

Explanation:

Let us recall that a negative ion is formed by addition of electrons to an atom. When electrons are added to the atom, greater interelectronic repulsion increases the size of the Te^2− hence it is greater in size than Te atom. Therefore, the ionic radius of Te^2− is greater than the atomic radius of Te.

In the second question, oxygen is positioned so far to the right because it has a far smaller nuclear charge compared to Te. Hence in the PES spectrum, the 1s sublevel of oxygen lies far to the right of that of Te.

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Copper has been used for thousands of years, either as a pure metal or in alloys. It is frequently used today in the production
motikmotik

The question is incomplete, the correct question is:

Copper has been used for thousands of years, either as a pure metal or in alloys. It is frequently used today in the production of wires and cables. Copper can be obtained through smelting or recycling. Determine the energy associated with each of these processes in order to recycle 1.08 mol Cu. The smelting of copper occurs by the balanced chemical equation: CuO(s)+CO(g) Cu(s)+CO2?(g) where ?H°f, CuO is = -155 kJ/mol. Assume the process of recycling copper is simplified to just the melting of the solid Cu starting at 25°C. The melting point of Cu is 1084.5°C with ?H°fus = 13.0 kJ/mol and a molar heat capacity, cp,Cu = 24.5 J/mol·°C.

Enthalpy change for the reaction recovering ?

Cu from CuO Energy for recycling Cu?

Answer:

Energy for recovering Cu from CuO = - 138. 24kJ

the total energy for recycling Cu is 42.07kJ

Explanation:

CuO(s) + CO(g) - - - - - - - > Cu(s) + CO2(g)

ΔHrxn = ΔHf(products) - ΔHf(reactants)

= ΔHf(CO2) - (ΔHf(CO)) + Δ Hf(CuO))

= - 393.5 kJ/mol - (-110.5 kJ/mol + ( - 155 kJ/mol)

= - 393.5 kJ/mol + 265.5 kJ/mol

= - 128 kJ/mol

for 1.08 mol of Cu

ΔH= - 128 kJ/mol × 1.08 mol = - 138. 24 kJ

Therefore,

Energy for recovering Cu from CuO = - 138. 24kJ

Part.2 :-

Total energy required = Heat required to raise the temperature of Cu from 25°C to 1084.5°C (q1) + Heat required to melt Cu at 1084.5°C(q2)

q1= n × ΔT × Cp

q1 = 1.08 mol × (1084.5°C - 25°C) × 24.5 J/mol 0C

q1 = 28.03 kJ

q2 = ΔHfus × n

q2 = 13.0 kJ/mol × 1.08 mol

q2 = 14.04kJ

Therefore,

Energy for recycling Cu = 28.03 kJ + 14.04kJ = 42.07kJ

Therefore, the total energy for recycling Cu is 42.07kJ

4 0
2 years ago
A paint machine dispenses dye into paint cans to create different shades of paint. The amount of dye dispensed into a can is kno
frosja888 [35]

Answer:

The 9th percentile is 4.464ml

Explanation:

Hello!

Given the variable X: Amount of dye dispensed into a paint can.

With a normal distribution, mean μ= 5ml and standard deviation δ= 0.4ml

You need to calculate the 9th percentile of the distribution.

The percentile is a measure of the position that indicates the number that separates the distribution or data set in a percentage of interest. In this case, the 9th percentile is the value of the distribution that separates the bottom 9% from the top 91%.

Symbolically:

P(X≤x₀)= 0.09

x₀ represents the value of the percentile.

The best way to calculate this value is by using the standard normal distribution since it is already tabulated, and then "translate" the Z-value to a value of the variable.

Under the standard normal distribution you have to look for:

P(Z≤z₀)= 0.09

The value marks the bottom 9% of the distribution, this means that you'll find it in the left tail of it. Remember the mean of the standard normal distribution is zero, so all values under the mean will be negative. Using the left entry of the Z-table you have to look for 0.09 in the body of the table and reach the margins to find the corresponding value:

z₀= -1.34

Now using the formula of the distribution you can "translate" the Z-value in terms of the variable of interest

Z=(X-μ)/ δ ~N(0;1)

z₀=(x₀-μ)/ δ

z₀*δ=x₀-μ

x₀=(z₀*δ)+μ

x₀=(-1.34*0.4)+5

x₀= 4.464

The 9th percentile is 4.464ml

I hope you have a SUPER day!

4 0
2 years ago
A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is
Alik [6]
<span>Answer: A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+ Kf for Cu(en)2^2+ is 1x10^20. so 1 Cu+2 & 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2 [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Cu+2 = 9.26 e-19 Molar since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
4 0
2 years ago
When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the result­ing concentration of Na+ is 2
tiny-mole [99]

Answer : The resulting concentration of Na^+ ion is, 4.5 M

Explanation : Given,

Concentration of Na_2CO_3 = M_1 = 3.0 M = 3.0 mol/L

Volume of Na_2CO_3 = V_1 = 70 mL = 0.07 L

Concentration of NaHCO_3 = M_2 = 1.0 M = 1.0 mol/L

Volume of NaHCO_3 = V_2 = 30 mL = 0.03 L

First we have to calculate the moles of Na_2CO_3 and NaHCO_3

\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol

and,

\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol

Now we have to calculate the moles of Na^+ ions.

As, 1 mole of Na_2CO_3 will give 2 moles of Na^+ ions

So, 0.21 moles of Na_2CO_3 will give 2\times 0.21=0.42 moles of Na^+ ions

and,

As, 1 mole of NaHCO_3 will give 1 mole of Na^+ ions

So, 0.03 moles of NaHCO_3 will give 0.03 moles of Na^+ ions

So,

Total number of moles of Na^+ ions = 0.42 + 0.03 =0.45 mole

Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L

Now we have to calculate the concentration of Na^+ ions.

\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M

Therefore, the resulting concentration of Na^+ ion is, 4.5 M

6 0
2 years ago
How is data not actually obtained from the experiment represented in a line graph? with a double line with only dots with a colo
Fittoniya [83]
The way how <span>data is not actually obtained from the experiment represented in a line graph is defnitely that </span><span>a colored line with a broken line. It is a well known fact that to obtain the actual data from the experiment you there should be plotted points on the line. Hope it will help you! Regards.</span>
7 0
2 years ago
Read 2 more answers
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