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anastassius [24]
2 years ago
6

If oxygen with an electronegativity of 3.5 were to make a chemical bond with lithium, electronegativity of 1.0, what kind of bon

d would form
Chemistry
1 answer:
sergejj [24]2 years ago
6 0

Answer:

Ionic.

Explanation:

Elements with higher electronegativity values are better at attracting electrons in a chemical bond.  

  • A chemical bond is considered "ionic" if the electronegativity difference between the two bonding atoms is greater than 1.8.
  • Otherwise, this chemical bond is considered "covalent".

In this example, the difference between the electronegativity of oxygen and lithium is 3.5 - 1.0 = 2.5. Since 2.5 > 1.8, the bond between the two elements would likely be ionic.

It is possible to reach the same conclusion based on the fact that lithium is a metal while oxygen is a nonmetal. When metal elements react with non-metal elements, the product is typically an ionic compound- with ionic bonds between the atoms.

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The reaction of hydrogen gas and iron oxide is described by the chemical equation below. 3H2+Fe2O3→2Fe+3H2O How many moles of ir
ELEN [110]

Answer:

2.2 moles of Fe will be produced

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 3.3 moles

Number of moles of iron oxide = 1.5 moles

Step 2: The balanced equation

3H2 + Fe2O3 → 2Fe + 3H2O

Step 3: Calculate the limiting reactant

For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O

Hydrogen gas is the limiting reactant. It will completely be consumed (3.3 moles). Fe2O3 is in excess. There will react 3.3 / 3 = 1.1 moles

There will remain 1.5 - 1.1 = 0.4 moles Fe2O3

Step 4: Calculate moles Fe

For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O

For 3.3 moles H2 we'll have 2/3 * 3.3 = 2.2 moles Fe

2.2 moles of Fe will be produced

5 0
3 years ago
The atom with the same number of neutrons as 12C is 9Be. 11B. 13C. 14N.
julia-pushkina [17]

Answer : The atom with the same number of neutrons as ^{12}C is, ^{11}B

Explanation :

  • The given atom is, ^{12}C

Atomic mass number = 12

Atomic number = 6

Atomic number = Number of protons = Number of electrons = 6

Number of neutrons = Atomic mass number - Atomic number = 12 - 6 = 6

Now we have to determine the number of neutrons in the given options.

  • In atom, ^{9}Be

Atomic mass number = 9

Atomic number = 4

Atomic number = Number of protons = Number of electrons = 4

Number of neutrons = Atomic mass number - Atomic number = 9 - 4 = 5

  • In atom, ^{11}B

Atomic mass number = 11

Atomic number = 5

Atomic number = Number of protons = Number of electrons = 5

Number of neutrons = Atomic mass number - Atomic number = 11 - 5 = 6

  • In atom, ^{13}C

Atomic mass number = 13

Atomic number = 6

Atomic number = Number of protons = Number of electrons = 6

Number of neutrons = Atomic mass number - Atomic number = 13 - 6 = 7

  • In atom, ^{14}N

Atomic mass number = 14

Atomic number = 7

Atomic number = Number of protons = Number of electrons = 7

Number of neutrons = Atomic mass number - Atomic number = 14 - 7 = 7

Therefore, the atom with the same number of neutrons as ^{12}C is, ^{11}B

6 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s) c) How many moles of the e
sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

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