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mart [117]
3 years ago
14

Bonds between metals and nonmetals tend to be ionic, but bonds between nonmetal atoms tend to be covalent. Explain why this is t

rue, using your understanding of electronegativity and ionization energy for these groups of elements.
Chemistry
2 answers:
Igoryamba3 years ago
5 0

Bond type depends on differences in electronegativity.


Metals and nonmetals tend to have large differences in electronegativity.


Nonmetals tend to have smaller differences in electronegativity.


Ionic bonds form between atoms that have large electronegativity differences.


Covalent bonds form between atoms with small electronegativity differences.

Assoli18 [71]3 years ago
3 0
Ionic bond is a type of bond in which one or more atoms are transferred to another atom which results to two ions with opposite charge. They attract each other.

The atom of another element is removed and then gained by another element which produces a noble gas electron configuration.

Covalent bond is a type of bond in which a pair or more electrons are being shared by the atoms of two elements.

Because of this, the molecules become stable by sharing the electrons thus creating a noble gas configuration for each of the atom.


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4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.
slava [35]

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ\frac{1mol}{102g} = <em>0,0980 moles</em>

And 10,0g of HCl are:

10,0 gₓ\frac{1mol}{36,5g} = <em>0,274 moles</em>

<em />

For a total reaction of 0,274 moles of HCl you need:

0,274×\frac{1molesAl_{2}O_3}{6 mole HCl} = <em>0,0457 moles of Al₂O₃</em>

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl ×\frac{2 moles AlCl_{3}}{6 moles HCl} × 133\frac{g}{mol} = <em>12,1 g of AlCl₃</em>

<em />

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles =<em> </em>0,0523 moles

And its mass is:

0,0523 molesₓ\frac{102g}{1mol} = <em>5,33 g of Al₂O₃ </em>

<em />

I hope it helps!

7 0
3 years ago
If wind is eroding an area at a rate of 2 mm per year and depositing it in a smaller area at a rate of 7 mm per year, how much l
amid [387]

Answer:

After 2000 years, the first after will be 4m lower.

After 2000 years, the second area will be 14m higher.

Explanation:

From the given information:

At the first area:

if the wind is eroding at 2 mm/year, then after 2000 years; we have:

2 mm/year × 2000 year = 4000 mm

4000 mm to meters, we have:

=4000 \ mm \times \dfrac{1 \ m}{ 1000 \ mm}

= 4 m lower

At the second smaller area:

Given that the wind is depositing it at the rate of 7 mm per year;

Then:

7 mm/year × 2000 year = 14000 mm

=14000 \ mm \times \dfrac{1 \ m}{ 1000 \ mm}

=14 m higher

6 0
3 years ago
Can anyone help with this? please
Sloan [31]

Answer:

Volume of solution = 80.5 mL

Explanation:

Given data:

Molarity  of solution = 4.50 mol/L

Mass of ethanol = 16.7 g

Volume of solution = ?

Solution:

Volume will be calculated from molarity formula.

Molarity  = number of moles / volume in L

Number of moles:

Number of moles = mass/molar mass

Number of moles = 16.7 g/ 46.07 g/mol

Number of moles = 0.3625 mol

Volume of solution:

Molarity  = number of moles / volume in L

4.50 mol/L = 0.3625 mol / volume in L

Volume in L =  0.3625 mol /4.50 mol/L

Volume in L = 0.0805 L

Volume in mL:

0.0805 L ×1000 mL/1 L

80.5 mL

3 0
3 years ago
Do you think this wing is from the left side or the right side of the chicken's body? Explain your answer.
lakkis [162]

Answer:

si

Explanation:

aveces

4 0
2 years ago
Consider the data [X] [Y] [Z] initial rate M M M M · s −1 Exp 1 0.30 0.20 0.35 0.210 Exp 2 0.60 0.10 0.70 0.420 Exp 3 0.60 0.20
sergij07 [2.7K]

Answer:

Rate = k [X]⁻¹ [Z]²

Explanation:

[X] [Y] [Z] initial rate M M M M · s −1

Exp 1 0.30 0.20 0.35      0.210

Exp 2 0.60 0.10 0.70      0.420

Exp 3 0.60 0.20 0.70     0.420

Exp 4 0.60 0.40 0.35      0.105

In Experiment 2 and 3 where the concentrations of Y and Z were constant, doubling the concentration of Y had no effect on the rate of the reaction. This means, that the rate of the reaction is zero order with respect to Y.

In experiment 3 and 4, dividing the concentration of Z by 2, causes the rate of the reaction to decrease by 4. This means the rate of the reaction is second order with respect to Z.

In experiment 1 and 4, doubling the concentration of X, causes the rate of the reaction to decrease by half. This means that X has an order of -1 with respect to the rate of the reaction.

The rate expression is given as;

Rate = k [X]⁻¹[Y]⁰[Z]²

Rate = k [X]⁻¹ [Z]²

3 0
3 years ago
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