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zhuklara [117]
3 years ago
13

When do Metallic bonds form?

Chemistry
1 answer:
Vanyuwa [196]3 years ago
3 0
Metallic bonds<span>, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize</span>
You might be interested in
A 50.6 grams sample of magnesium hydroxide (Mg(OH)2) is reacted with 45.0 grams of hydrochloric acid (HCl). What is the theoreti
zysi [14]
            <span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)

grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required

Since there are only 45.0 grams HCl, then HCl is the limiting reactant.

theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
7 0
3 years ago
Read 2 more answers
When forming a chemical bond only the ________ electrons participate in the bonding process.
matrenka [14]
The outer shell electrons are only involved in the bonding process since they are the only 'incomplete' shell and it needs to be fulfilled by another element. 
8 0
3 years ago
Read 2 more answers
Find the percent composition for SF6<br> S =? %<br> F =? %
goldenfox [79]

Answer:

S = 21.92 %

F = 78.08 %

Explanation:

To find the percent composition of each element in SF6, we must find the molar mass of SF6 first.

Molar mass of SF6 = 32 + 19(6)

= 32 + 114

= 146g/mol

mass of Sulphur (S) in SF6 = 32g

mass of Fluorine (F) in SF6 = 114g

Percent composition = mass of element/molar mass of compound × 100

- % composition of S = 32/146 × 100 = 21.92%.

- % composition of F = 114/146 × 100 = 78.08%.

6 0
3 years ago
Mg + 2HCl → MgCl2 + H2
deff fn [24]

Answer:

Explanation:

This is a limiting reactant problem.

Mg(s)

+

2HCl(aq)

→

MgCl

2

(

aq

)

+ H

2

(

g

)

Determine Moles of Magnesium

Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).

4.86

g Mg

×

1

mol Mg

24.3050

g Mg

=

0.200 mol Mg

Determine Moles of 2M Hydrochloric Acid

Convert

100 cm

3

to

100 mL

and then to

0.1 L

.

1 dm

3

=

1 L

Convert

2.00 mol/dm

3

to

2.00 mol/L

Multiply

0.1

L

times

2.00 mol/L

.

100

cm

3

×

1

mL

1

cm

3

×

1

L

1000

mL

=

0.1 L HCl

2.00 mol/dm

3

=

2.00 mol/L

0.1

L

×

2.00

mol

1

L

=

0.200 mol HCl

Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,

2.01588 g/mol

0.200

mol Mg

×

1

mol H

2

1

mol Mg

×

2.01588

g H

2

1

mol H

2

=

0.403 g H

2

0.200

mol HCl

×

1

mol H

2

2

mol HCl

×

2.01588

g H

2

1

mol H

2

=

0.202 g H

2

The limiting reactant is

HCl

, which will produce

0.202 g H

2

under the stated conditions.

pls mark as brainliest ans

7 0
2 years ago
Read 2 more answers
assuming nitrogen behaves like an ideal gas, what volume would 14.0 g of nitrogen gas (N2) occupy at STP? the gas constant is 0.
dimaraw [331]

Answer:

V = 22.41 L

Explanation:

Given data:

Mass of nitrogen = 14.0 g

Volume of gas at STP = ?

Gas constant = 0.0821 atm.L/mol.K

Solution:

Number of moles of gas:

Number of moles = mass/molar mass

Number of moles= 14 g/ 14 g/mol

Number of moles = 1 mol

Volume of gas:

PV = nRT

1 atm × V = 1 mol × 0.0821 atm.L/mol.K  × 273 K

V = 22.41 atm.L / 1 atm

V = 22.41 L

4 0
3 years ago
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