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ratelena [41]
3 years ago
5

List the following bond types in order of increasing strength: non-polar covalent bonds, ionic bonds, hydrogen bonds, polar cova

lent bonds.
A. Hydrogen bonds, ionic bonds, non-polar covalent bonds,​ polar covalent bonds.
B. Hydrogen bonds, non-polar covalent bonds,​ polar covalent bonds, ionic bonds.
C. Hydrogen bonds, ionic​ bonds, polar covalent​ bonds, non-polar covalent​ bonds.
D. Ionic bonds, polar covalent bonds, non-polar​ covalent​ bonds, Hydrogen bonds
E. Non-polar covalent bonds, polar covalent​ bonds, Hydrogen bonds, ionic bonds.
Chemistry
1 answer:
ss7ja [257]3 years ago
8 0

Answer: Option (B) is the correct answer.

Explanation:

  • An ionic bond is formed by the sharing of electrons between two chemically combining atoms.

In an ionic bond, there occurs attraction between oppositely charged ions due to which there occurs strong forces of attraction between them. Therefore, ionic bonds are the strongest bonds.

  • A polar covalent bond is formed due to unequal sharing of electrons between the combining atoms.

For example, H_{2}O is a polar covalent compound. Partial opposite charges tend to develop on the atoms of a polar covalent compound.

  • A non-polar covalent bond is formed due to equal sharing of electrons between the combining atoms.

For example, Cl_{2} is a non-polar covalent molecule. No partial charges will be there on the atoms of a non-polar covalent molecule.

  • A hydrogen bond is defined as the bond formed between a hydrogen atom and an electronegative atom.

For example, in HCl compound there occurs hydrogen bonding.

In this type of bond, dipole-dipole attractive interactions tend to take place. And, strength of hydrogen bonds is very weak.

Thus, we can conclude that given bond types are arranged in order of increasing strength as follows.

      Hydrogen bonds < non-polar covalent bonds < polar covalent bonds < ionic bonds

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Identify the products in the reaction HCl+NaOH=NaCl+H2O
valentina_108 [34]

NaCl and H2O.

The products are typically the elements/compounds on the right side of the equation or the right side of the arrow. The left side of the arrow would be the reactants of the equation.

Hope this helps!

5 0
3 years ago
PLEASE HELP ME ASAP! CHEMISTRY TUTOR<br><br> SEE ATTACHED
Masteriza [31]

Answer:

\large \boxed{\text{-827.4 kJ}}

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + <u>2H₂O(g</u>)  ⟶ 2PbO(s) + <u>2H₂S(g</u>);  ∆H =  208.6 kJ

6. <u>2H₂S(g)</u> + O₂(g)        ⟶ <u>2S(s</u>)     + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ

<u>7</u><u>. </u><u>2S(s)</u><u>      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ </u>

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}

8 0
3 years ago
The mole is defined as the amount of a substance containing the same number of particles as exactly 12 g of C-12. The amu is def
Pavlova-9 [17]

Answer:

Because it wouldn't make any sense

Explanation:

First of all, I think it's important to highlight the definition of isotope.

Isotope (Wikipedia): Variants of a particular chemical element <u>which differ in neutron number</u>...

This means that two isotopes are the same element but have different net mass per atom, due to the different number of neutrons.

Therefore, it's important to make the definition on the same isotope so that the proportion is equal. If the definition would be made on different isotopes, the proportion wouldn't have any sense. Let me be clear with this example:

mass in grams of a C-12 atom = 1.9944235 × 10 ^ -23 g  --> this is the mass of a single C atom.

By definition --> 1 mol of anything = 6.02 x 10 ^ 23 particles of anything

Therefore, we know how much a single C atom weights. How many grams do you think that 6.02 x 10 ^ 23 atoms of C (i.e a mol of C) could weight??

1 single C atom ----------------------------- 1.9944235 × 10 ^ -23 g

6.02 x 10 ^ 23 atoms of C   ------------ <u>12.006 g  !!</u>

These 12 g is the same quantity than above! Therefore, 1 mol of C weights 12 g. If the definition were made with 13 g of C-13 (the other C isotope), these numbers will not be the same --> There would be a contradiction.

Regarding the second question, we need to search Ne-20 atomic mass in grams -->  3,3509177 × 10 ^ -23 g

Hence, if we follow the same rule, the amu would be 1/12 of Ne-20.

[ 3,3509177 × 10 ^ -23 g ] / 12 = 2.79 ^ -24 g

3 0
3 years ago
Repeating many forms a crystal structure inside each of these repeated units is the ?
djyliett [7]

Answer:

gggzbjv

Explanation:

vhbbn

nnkjj

vjhgnkj

jjjjjjj

7 0
2 years ago
Read 2 more answers
The Greek letter delta (triangle sign) above the arrow in a chemical equation means.
kogti [31]
“yields” or “produces”
3 0
3 years ago
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