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dusya [7]
2 years ago
7

The properties of two elements are listed below. Element Atomic radius (pm) Ionic radius (pm) First ionization energy (kJ/mol) E

lectron affinity (kJ/mol) Electronegativity Br 114 195 1140 –325 3.0 K 243 152.2 418.8 –48.4 0.82 Which prediction is supported by the information in the table? K will give up an electron more easily than Br. Both K and Br will have the same pull on electrons. K will have a smaller size in comparison to Br. Both K and Br will produce ions of the same size.
Chemistry
2 answers:
kifflom [539]2 years ago
5 0

Based on the data presented in the table and comparing it to the given statements we can interpret that potassium (k) will give up electrons readily than bromine (Br)

Now, ionization energy is the amount of energy required to remove an electron from an atom or ion. In the case of K the first IE = 418.8 kJ/mol while for Br, IE = 1140 kj/mol. Thus the ionization energy of Br >> K which supports the prediction that K will give up an electron more easily than Br.

Helen [10]2 years ago
3 0
The First option may be correct. Im deeply sorry if im wrong!

FIRST OPTION (A)(1)
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A metal, M , of atomic mass 56 amu reacts with chlorine to form a salt that can be represented as MClx. A boiling point elevatio
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Answer:

  MCl₂

Explanation:

The formula for boiling point elevation can be used to find x. The "complete dissociation" means there will be an ion of M and x ions of Cl in the solution. The number of moles of solute will be 30.2 grams divided by the molecular weight of MClx, where x is the variable we're trying to find.

  \Delta T=imK_b\qquad\text{where i=ions/mole, m=molality, $K_b\approx 0.512$}\\\\376.81-373.15=(x+1)\dfrac{\text{moles}}{\text{kg solvent}}(0.512)\\\\\dfrac{3.66}{0.512}=(x+1)\dfrac{\dfrac{30.2}{56+35.45x}}{0.1}=\dfrac{302(x+1)}{56+35.45x}\\\\\dfrac{3.66}{0.512\cdot 302}(56+35.45x)=x+1\\\\\dfrac{3.66\cdot 56}{0.512\cdot 302}-1=x\left(1-\dfrac{3.66\cdot 35.45}{0.512\cdot 302}\right)\\\\x=\dfrac{50.336}{24.877}\approx 2.023

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6 0
2 years ago
If South America were not there, explain how the direction of South equatorial current would be different
forsale [732]

Explanation:

South America land mass serves as a deflector for the South equatorial current. This deflection the current causes them to move in a different direction. If the Continent were not present the direction of the South equatorial current would not change and it would continue to flow in the west.

7 0
2 years ago
Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM
7nadin3 [17]

Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will  B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: <u><em>The value of Kc remains the same.</em></u>

<u><em> </em></u>As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, <u><em>The reaction must: A. run in the forward direction to reestablish equilibrium</em></u>.

By decreasing the volume, and so that Kc remains constant, being:

Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2}  ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume}  } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume

 where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so,  the number of moles of Cl₂ should decrease.<u><em>The number of moles of Cl2 will  B. decrease.</em></u>

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, <em><u>the value of Qc C. is less than Kc.</u></em>

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