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Nitella [24]
3 years ago
5

Four atoms and/or ions are sketched below in accordance with their relative atomic and/or ionic radii. Which of the following se

ts of species are compatible with the sketch?
Explain. (a) C,Ca2+,Cl−,Br−;
(b) Sr4, Cl,Br−,Na+

(d) Al,Ra2+,Zr2+

(c) Y,K,Ca,Na+, Mg2+;

e) Fe,Rb,Co,Cs
Chemistry
1 answer:
kotykmax [81]3 years ago
4 0

Answer:

Hence the correct option is an option (b) Sr4, Cl,Br−,Na+.

Explanation:    

Bromine and chlorine belong to an equivalent group. As we go down the group the dimensions increases which too there's a charge on the bromine atom. therefore the size of the Br- is going to be larger in comparison to the chlorine atom.

Sr atom is within the second group, and also it's below the above-mentioned atoms.so Sr is going to be the larger one among all the atoms.

Sodium and chlorine belong to an equivalent period .size decrease from left to right. but due to the charge on sodium its size decreases and there's an opportunity that Na+ size could be adequate for Cl.      

Here we finally assume that two atoms are of an equivalent size (Na+ and Cl) which are less in size compared to the opposite two(Sr and Br-) during which one is greater (Sr)and the opposite is smaller(Br-).

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Predict the sign of the entropy change of the system for each of the following reactions.
storchak [24]

Answer :

(a) The entropy change will also decreases.

(b) The entropy change will also increases.

(c) The entropy change will also decreases.

(d) The entropy change will also increases.

Explanation :

Entropy : It is defined as the measurement of randomness or disorderedness in a system.

The order of entropy will be,

As we are moving from solid state to liquid state to gaseous state, the entropy will be increases due to the increase in the disorderedness.

As we are moving from gaseous state to liquid state to solid state, the entropy will be decreases due to the decrease in the disorderedness.

(a) N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy change will also decreases.

(b) CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

In this reaction, the randomness of reactant molecules are less and as we move towards the formation of product the randomness become more that means the degree of disorderedness increase. So, the entropy change will also increases.

(c) 3C_2H_2(g)\rightarrow C_6H_6(g)

In this reaction, 3 mole of gaseous C_2H_2 react to give 1 mole of gaseous C_6H_6 that means randomness become less that means the degree of disorderedness decreases. So, the entropy change will also decreases.

(d) Al_2O_3(s)+3H_2(g)\rightarrow 2Al(s)+3H_2O(g)

In this reaction, the number of moles of gases are same on both side of the reaction but 1 mole of solid Al_2O_3 react to give 2 moles of solid aluminium that means randomness become more that means the degree of disorderedness increases. So, the entropy change will also increases.

5 0
3 years ago
Please help as soon as possible, Thank you!
strojnjashka [21]

Answer:

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Explanation: warm winters and cool summers

4 0
3 years ago
The noble gases are located in which column of the periodic table?A. 1B. 6C. 9D. 18
Vika [28.1K]
D IS THE ANSWER TO YOUR QUESTION
6 0
3 years ago
What is the chemical formula for the ionic compound calcium chloride?
gulaghasi [49]

C a C l ^2


Hope this helps

7 0
3 years ago
Read 2 more answers
Consider the halogenation of ethene, where x is a generic halogen: h2c=ch2(g)+x2(g)→h2xc−ch2x(g) you may want to reference (page
KengaRu [80]
Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a) 
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ

2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ

2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ

2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ

Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction
8 0
3 years ago
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