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balu736 [363]
4 years ago
9

Identify the compound with the highest magnitude of lattice energy. Identify the compound with the highest magnitude of lattice

energy. CaBr2 CaBr2 MgBr2 SrBr2
Chemistry
1 answer:
Galina-37 [17]4 years ago
4 0

Answer: Option (c) is the correct answer.

Explanation:

According to the Born-Lande equation,

          E = \frac{N_{A}Mz^{+}z^{-}e^{2}}{4 \pi \epsilin_{o} r}

where,   N_{A} = Avogadro's constant

                    r = distance between the anion and cation

                   M = Medlung constant

           z^{+} = charge on the cation

           z^{-} = charge on the anion

                  E = lattice energy

According to this expression, lattice energy is inversely proportional to the distance between the cation and anion. And, when we move down a group then there occurs an increase in the atomic radii of the atoms.

As a result, there will occur a decrease in their lattice energy. Since, the anionic part is same in all the given species but the cationic part is different.

In MgBr_{2} the size of magnesium cation is smaller than calcium, barium and strontium ion. Hence, there will be least distance between Mg and Br ions.

Thus, we can conclude that the magnitude of lattice energy will be the highest in MgBr_{2}.

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How many moles are in 1.5 x 10^20 molecules of NH3?
Keith_Richards [23]
<span>Because the question is asking moles of NH3, the compound, any subscripts are irrelevant. It only wants to know how many moles of NH3 molecules, not individual atoms.

Therefore, we can simply convert to moles. 1.5x10^23/6.022x10^23 = .249 moles of NH3.

(If it were to ask moles of Hydrogen, for example, you would multiply the answer by 3, because there are 3 atoms of Hydrogen per one molecule of NH3. But this only asks for moles of the entire compound).  

hope you have a great day! :)
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5 0
4 years ago
What is the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml?
Firdavs [7]

Answer: The final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,  

M_1 = molarity of stock CuSO_4 solution = 2.5 M

V_1 = volume of stock CuSO_4solution = 5 ml  

M_1 = molarity of diluted CuSO_4 solution = ?

V_1 = volume of diluted CuSO_4 solution = 750 ml

Putting in the values we get:

2.5\times 5=M_2\times 750

M_2=0.017

Therefore the final concentration when 5 ml of a 2.5M copper sulphate solution is diluted to 750 ml is 0.017 M

6 0
3 years ago
What is the energy of a photon with the frequency of 1.02 x 1015 Hz?<br> Please show your work!
galben [10]
<span> E = h v where v = frequency h = plancks constant 6.626 * 10 - 34

4.8 * 6.626 = 31.8 </span>
7 0
3 years ago
HELP ASAP PLEASE!!!! 15 PTS!!
Bond [772]
ADD THEM all, and then divide by four. Thats what I would do! 
5 0
3 years ago
Read 2 more answers
Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o
stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0
3 years ago
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