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horrorfan [7]
3 years ago
11

How to determine if the ionic bond between lithium fluoride is stronger or weaker than the ionic bond in magnesium oxide?

Chemistry
1 answer:
lara [203]3 years ago
4 0

Explanation:

As we know that the atomic number of lithium is 3 and its electronic distribution is 2, 1. Whereas atomic number of fluorine is 9 and its electronic distribution is 2, 7. So, for a metal it is easier to lose an electron rather than losing 2 or more number of electrons.

Also, fluorine is much more electronegative in nature. Hence, it will readily accept an electron from a donor atom. Therefore, a strong ionic bond will be present between lithium and fluorine atom.

On the other hand, atomic number of magnesium is 12 and its electronic configuration is 2, 8, 2. And, atomic number of oxygen is 8 and its electronic distribution is 2, 6. Also, oxygen is less electronegative than fluorine therefore, force of attraction exerted by oxygen to gain the valence electrons will not be strong enough.

Hence, the ionic bond between magnesium oxide is not strong enough.  

Thus, we can conclude that ionic bond between lithium fluoride is stronger than ionic bond in magnesium oxide.

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How many moles of hydrogen are produced from 57.5 grams of lithium?
melomori [17]

The moles of hydrogen which are produced from 57.5 grams of lithium is 8.21 grams.

<h3>What is stoichiometry?</h3>

Stoichiometry of the reaction gives idea about the relative amount of product produced by the reactant in terms of moles.

Given chemical reaction is:

2Li(s) + H₂O(l) → 2LiOH(aq) + H₂(g)

Moles of lithium will be calculated as:

n = W/M, where

  • W = given mass = 57.5g
  • M = molar mass = 7g/mol

n = 57.5 / 7 = 8.21 mol

From the stoichiometry of the reaction:

  • 2 moles of Li = produces 1 mole of H₂
  • 8.21 moles of Li = produces 8.21/2=4.105 moles of H₂

Mass of H₂ = (4.105mol)(2g/mol) = 8.21g

Hence required mass of hydrogen gas is 8.21g.

To know more about mass & moles, visit the below link:

brainly.com/question/20563088

#SPJ1

8 0
2 years ago
Manganese(II) oxide, lead(IV) oxide, and nitric acid react to produce permanganic acid, lead(II) nitrate, and water according to
Leokris [45]

Answer:

There will be produced:

2.97 moles HMnO4

4.45 moles Pb(NO3)2

2.97 moles H2O

Explanation:

Step 1: Data given

Manganese(II) oxide = MnO2

lead(IV) oxide = PbO2

nitric acid = HNO3

Moles of HNO3 = 8.90 moles

Step 2: The balanced equation

2MnO2 + 3PbO2 + 6HNO3 → 2HMnO4 + 3Pb(NO3)2 + 2H2O

Step 3: Calculate moles of reactants and products

For 2 moles MnO2 we need 3 moles PbO2 and 6 moles HNO3 to produce 2 moles HMnO4, 3 moles Pb(NO3)2 and 2 moles of water

For 8.90 moles of HNO3, there will react:

8.90 / 3 = 2.97 moles MnO2

8.90 / 2 = 4.45 moles PbO2

There will be produced:

8.90/3 = 2.97 moles HMnO4

8.90/2 = 4.45 moles Pb(NO3)2

8.90 / 3 = 2.97 moles H2O

7 0
3 years ago
What is density?
kirza4 [7]
Answer: 3, the amount per unit volume of a particular material
7 0
3 years ago
Read 2 more answers
Determine the mass of water produced from 50g hydrochloric acid and
Alenkinab [10]

Answer:

24.7 grams H₂O

Explanation:

To find the mass of water, you should (1) convert from grams HCl to moles (via molar mass from periodic table), then (2) convert from moles HCl to moles H₂O (via mole-to-mole ratio from equation), and then (3) convert from moles H₂O to grams (via molar mass from periodic table).

1 HCl + KOH --> 1 H₂O + KCl

Molar Mass (HCl): 1.008 g/mol + 35.45 g/mol
Molar Mass (HCl): 36.458 g/mol

Molar Mass (H₂O): 2(1.008 g/mol) + 16.00 g/mol
Molar Mass (H₂O): 18.016 g/mol

50 g HCl        1 mole HCl          1 mole H₂O           18.016 g
--------------  x  ------------------  x  -------------------  x  -------------------  = 24.7 g H₂O
                        36.458 g           1 mole HCl          1 mole H₂O

3 0
2 years ago
In one experiment, the reaction of 1.00 g mercury and an excess of sulfur yielded 1.16 g of a sulfide of mercury as the sole pro
Ksju [112]

Based on experiment 1:

Mass of Hg = 1.00 g

Mass of sulfide = 1.16 g

Mass of sulfur = 1.16 - 1.00 = 0.16 g

# moles of Hg = 1 g/200 gmol-1 = 0.005 moles

# moles of S = 0.16/32 gmol-1 = 0.005 moles

The Hg:S ratio is 1:1, hence the sulfide is HgS

Based on experiment 2:

Mass of Hg taken = 1.56 g

# moles of Hg = 1.56/200 = 0.0078

Mass of S taken = 1.02 g

# moles of S = 1.02/32 = 0.0319

Hence the limiting reagent is Hg

# moles of Hg reacted = # moles of HgS formed = 0.0078 moles

Molar mass of HgS = 232 g/mol

Therefore, mass of HgS formed = 0.0078 * 232 = 1.809 g = 1.81 g

3 0
3 years ago
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