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cluponka [151]
3 years ago
6

Predict which of the following pair will have the higher melting point and indicate why: KBr, Br2.

Chemistry
1 answer:
allsm [11]3 years ago
8 0

Answer:

KBr: strong ion-ion forces

Explanation:

Ionic compounds are non-molecular species. During melting they require a lot of energy input so that the strong ionic bonds that constitute the electrostatic lattice be disrupted. It thus requires very high temperatures. This is indicative of a very strong ion-ion electrostatic interaction which is typical of ionic bonds and results in high melting points of ionic solids.

Hence, the process of melting an ionic solid needs the addition of a large amount of energy in order to break all of the ionic bonds in the crystal.

On the other hand, Br2 is a covalent molecular specie. Most covalent compound have low melting points since their molecules are mostly held together only by weak van der waals forces.

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Zinc metal can be obtained from Zinc Oxide by reaction at high temperature with CO. The CO is obtained from burning C in limited
bonufazy [111]

Answer:

(a): 2ZnO(s) -----------> 2Zn(s) + O2(g)

2C(s) + O2(g) ---------> 2CO(g)

(b): ZnO(s) + C(s) ---------> Zn(s) + CO(g)

3 0
3 years ago
Dissolving brass requires an oxidizing acid such as concentrated nitric acid. Nitrogen dioxide is produced as a byproduct in thi
polet [3.4K]

Answer:

                  Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  2 NO₂  +  2 H₂O

Explanation:

Step 1: Write down the chemical formulas of given substances,

                                    Copper Metal  =  Cu

                                    Nitric Acid  =  HNO₃

                                    Copper (II) Nitrate  =  Cu(NO₃)₂

                                    Nitrogen Dioxide  =  NO₂

                                    Water  =  H₂O

Step 2: Write down the unbalance Chemical equation,

                         Cu  +  HNO₃    →    Cu(NO₃)₂  +  NO₂  +  H₂O

Step 3: Balance Cu atoms on both sides;

The number of Cu atoms on both sides are same. Hence, there number will remain the same.

Step 4: Balance N atoms on both sides;

As there is 1 N atom on left hand side and 3 N atoms on right hand side, so we will multiply HNO₃ by 3 to balance N on both sides, hence,

                         Cu  +  3 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  H₂O

Step 5: Balance O atoms on both sides;

As there are 9 O atom on left hand side and 9 O atoms on right hand side, so they are balance.

Step 6: Balance H atoms on both sides;

As there are 3 H atom on left hand side and 2 H atoms on right hand side, so we will multiply H₂O by 2 as,

                         Cu  +  3 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  2 H₂O

By doing so the number of O atoms got imbalanced, so to balance O atoms again we will multiply HNO₃ by 4 as,

                         Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  2 H₂O

Now, The Cu and H atoms are balanced, and the O atoms are greater on left hand side and the N atoms are greater on right hand side, therefore we will multiply NO₂ by 2 to balance both N and O as,

                         Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  2 NO₂  +  2 H₂O

7 0
3 years ago
What is the purpose of graphics in scientific articles
kow [346]

Answer:

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4 0
3 years ago
HELPPPPPPPPP ASAPPPP!!!!!
andreev551 [17]
Answer :
B

!!!!!!!!!!!!!!!
8 0
3 years ago
A polar covalent bond will form between which two atoms?
Alex_Xolod [135]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Be and F,

                    E.N of Fluorine          =   3.98

                    E.N of Beryllium        =   1.57

                                                   ________

                    E.N Difference                2.41          (Ionic Bond)


For H and Cl,

                    E.N of Chorine           =   3.16

                    E.N of Hydrogen        =   2.20

                                                   ________

                    E.N Difference                0.96          (Polar Covalent Bond)


For Na and O,

                    E.N of Oxygen          =   3.44

                    E.N of Sodium          =   0.93

                                                       ________

                    E.N Difference                2.51          (Ionic Bond)


For F and F,

                    E.N of Fluorine          =   3.98

                    E.N of Fluorine          =   3.98

                                                        ________

                    E.N Difference                0.00         (Non-Polar Covalent Bond)

Result:

           A polar covalent bond is formed between Hydrogen and Chlorine atoms.

5 0
3 years ago
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