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cluponka [151]
3 years ago
13

Do you expect the ionization energies of anions of group 17 elements to be lower than, higher than, or about the same for neutra

l atoms of the same group?
Chemistry
1 answer:
Minchanka [31]3 years ago
4 0

Answer: The ionization energies of anions of group 17 elements will be higher than for neutral atoms of the same group.

Explanation:

The ionization energy is the energy required to remove an electron from the valence shell of an isolated gaseous atom of an element.

Ionization energy depends on:

a) Electronic configuration of an element: If on losing or gaining of an electron an element attains noble gas configuration or half filled stability then it will be difficult to remove the electron from stable configuration.

b) size of an element :larger the size more lesser the effective nuclear charge on electron thus lower the ionization energy.

Group 17 elements have 7 valence electrons and thus on gaining one electron , they attain stable noble gas configuration and thus the ionization energy will become higher.

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A 100.0 ml sample of 0.10 m ca(oh)2 is titrated with 0.10 m hbr. determine the ph of the solution after the addition of 400.0 ml
WINSTONCH [101]
The balanced equation for the reaction is as follows;
Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O
stoichiometry of Ca(OH)₂ to HBr is 1:2
number of Ca(OH)₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol
Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol 
1 mol of Ca(OH)₂ needs 2 mol of HBr for neutralisation
therefore 0.010 mol of Ca(OH)₂  needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised
but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol 
then pH of the medium can be calculated using the excess H⁺ ions
HBr is a strong acid therefore complete ionization
[HBr] = [H⁺]
[H⁺] = 0.020 mol / (100.0 + 400.0 mL)
      = 0.020 mol / 0.5 L 
      = 0.040 mol/L
pH = -log[H⁺] 
pH = - log [0.040 M]
pH = 1.40
pH of the medium is 1.40
8 0
3 years ago
For a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k, calculate the inner diameter if you are designing for
bonufazy [111]

The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m

<h3>What is Stack Height ?</h3>

Stack height means the distance from the ground-level elevation at the base of the stack to the crown of the stack.

If a stack arises from a building or other structure, the ground-level elevation of that building or structure will be used as the base elevation of the stack.

Given is a steel stack that exhausts 1,200 cu.m/min of gases

P= 1 atm and

T= 400 K

maximum expected wind speed at stack height of 12 m/s

The formula for the diameter of chimney

\rm d=\sqrt{\dfrac{4Q}{\pi v} }

Q =1200 cu.m/min

= 1200 * 0.0166 = 19.92 cu.m/sec

Velocity = 12m/s

\rm d=\sqrt{\dfrac{4\times 19.92}{3.14*12} }\\

d= 1.45 m

Therefore The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m.

To know more about Stack Height

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8 0
2 years ago
I need help with chemistry :(
Yanka [14]

Answer:

How may we help kind sir

Explanation:

and if this was for points thanks

4 0
2 years ago
Read 2 more answers
A chemical equation is balanced when the
Arada [10]

a balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.

7 0
3 years ago
Read 2 more answers
Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalen
alexgriva [62]

Answer:

V_{base}=24.04mL

Explanation:

Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O

We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

2M_{base}V_{base}=M_{acid}V_{acid}

Therefore, for is to compute the volume of the used base, we proceed as shown below:

V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}

And we plug in to obtain:

V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL

Best regards!

8 0
3 years ago
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