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Aleksandr [31]
3 years ago
15

Which element combination(s) would result in electrons being shared?

Chemistry
2 answers:
Likurg_2 [28]3 years ago
5 0
Correct option is c) oxygen and hydrogen.... like in <u><em>H2O .. </em></u><em> </em> there is covalent bond due to sharing of electrons ...<em />
Olin [163]3 years ago
5 0

Explanation:

Atoms that share electrons with each other tend to form a covalent bond. Whereas atoms that show transfer of electrons from one atom to another tend to form an ionic bond.

Sodium has atomic number 11 and oxygen has atomic number 8. In order to become stabilized, sodium needs to lose one electron whereas oxygen needs to gain two electrons. So, two sodium atoms will combine with one atom of oxygen. This will form an ionic bond between sodium and oxygen atom.

Carbon has atomic number 4 and fluorine has atomic number 9. Therefore, both of them will share electrons in order to become stabilized. Thus, they will form a covalent bond.

Oxygen atom has atomic number 8 and hydrogen has atomic number 1. Therefore, both of them will share electrons but due to difference in electronegativity of both hydrogen and oxygen they will result in the formation of a polar covalent bond.

Aluminium has atomic number 13 and chlorine has atomic number 17. So, aluminium will donate or transfer its valence electron to chlorine atom. Therefore, it will also form an ionic bond.

Thus, we can conclude that carbon (C) and fluorine (F) and oxygen (O) and hydrogen (H) are the combination(s) that would result in sharing of electrons.

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MariettaO [177]

1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Explanation:

The formula used in solving the problems is

number of moles= \frac{mass}{atomic mass of one mole}      1st equation

molarity = \frac{number of moles}{volume}            2nd equation

Dilution formula

M1V1 = M2V2          3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n = \frac{21.6}{68.946}

  = 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume = \frac{0.313}{1.3}

            = 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles = \frac{215.1}{36.46}

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M = \frac{5.899}{2}

   = 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

  M1= 12 M

  V2=?

  M2= 4

applying the equation 3

50 x 12 = 4 x v2

V2 = 150 ml.

6. data given:

HCl + NaOH ⇒ NaCl + H20

molarity of NaOH = 0.525 M

volume of NaOH = 25 ml

molarity of acid HCl= 75 ml

volume of HCl = 0.335 ml

pH=?

Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119  moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity = \frac{0.0119}{0.1}

             = 0.11 M (pOH Concentration)

14 = pH + pOH  

  pH  = 14 - 0.11

     pH    = 13.89

3 0
3 years ago
The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14. Show all calculations lead
stich3 [128]
Kw = 1.219x10⁻¹⁴

kw = [H⁺] x [OH⁻]

in water [H⁺] = [OH⁻] therefore:

[H⁺] = √ kw

[H⁺] = √ <span>1.219x10⁻¹⁴

[H</span>⁺] = 1.104x10⁻⁷

pH = - log [H⁺]

pH = - log [ <span>1.104x10⁻⁷ ]

pH = 6.95

hope this helps!</span>
8 0
3 years ago
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