Who can help me with this problem Rank the following elements by electron affinity from most positive to most negative I have the elements Ar Cl O Sb and Rb?

Who can help me with this problem Rank the following elements by electron affinity from most positive to most negative I have the elements Ar Cl O Sb and Rb?
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Answer :

The order is: O, Cl, Sb, Rb, Ar.
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From the elements: Cl, Br, F, O, Al, C, Li, Cs and Xe; choose the following: (a) The element with highest negative electron gain enthalpy. (b) The element with lowest ionization enthalpy. (c) The element with smallest ionic radius. (d) The element with six electrons in the valence shell. (e) The element which is a liquid at room temperature. (f)The element which belongs to zero group. (g) The elements which forms largest number of compounds.

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Last Answer : Ans. (a). The element chlorine (Cl) has the highest negative electron gain enthalpy. (b). The element cesium (Cs) has lowest ionization enthalpy. (c). The element fluorine F has lowest atomic radius. (d ... zero group (or group 18). (g). The element carbon (C) forms the largest number of compounds.

The increasing order of reactivity among group 1 elements is Li<Na<K<Rb<Cs whereas among the group 17 elements, it is F>Cl>Br>I. Explain.

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Arrange the following order of the property indicated: (a). F, Cl, Br and I (negative electron gain enthalpy) (b). Mg, Al, Si and Na (ionization enthalpy) (c). C, N, O and F (second ionization enthalpy)

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Which out of F or Cl has a more negative electron gain enthalpy?

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Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

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Among the following, the number of elements showing only one non-zero oxidation state is: O,Cl,F,N,P,Sn,Tl,Na,Ti

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The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

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Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

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Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

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Consider the following reaction : `xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+) to xMn^(2+)+2yCO_(2)+(z)/(2)H_(2)O` The value of x, y and z in the reaction ar

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Last Answer : Consider the following reaction : `xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+) to xMn^(2+)+2yCO_(2)+(z)/(2)H_(2) ... 2, 5 and 8 C. 2, 5 and 16 D. 5, 2 and 8

Consider the following reaction : `xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+) to xMn^(2+)+2yCO_(2)+(z)/(2)H_(2)O` The value of x, y and z in the reaction ar

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The first forming microbial elements of plaque are: A. Aerobic gram positive G+ B. Aerobic gram negative G C. Anaerobic gram negative G D. Spirochetes E. Anaerobic gram positive G+

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Can blood group o positive donate a kidney to o negative?

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Can someone who has the blood type B negative donate blood to someone who is O positive?

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Can o positive blood be given to o negative patient?

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