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maria [59]
3 years ago
13

Which pair of atoms would form a covalent bond ?

Chemistry
2 answers:
Liula [17]3 years ago
7 0
I think O and Cl will form covalent bonds since they are both non-metals and don't have a large enough difference in electronegativities to create an ionic bond.

I hope this helps.  Let me know if anything is unclear.
NeTakaya3 years ago
4 0
<h2>Answer:</h2>

oxygen (O) and chlorine (Cl).

<h2>Explanation:</h2>

As the electronegativity value of an element increases It is more likely to make an ionic bond. e.g. the metallic elements. On the moderate values of electronegativity, elements tend to make covalent bond.

Oxygen and chlorine will make covalent bond because the electronegativity difference between these two elements is 0,5 so there bond will be covalent in nature.

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If a aluminum rod was made out of solid glass instead of solid aluminum what would happen.
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it would break when it hits the ground

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In Period 2, as the elements are considered from left to
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A substance that produces hydroxide ions when placed in water is a(n) _____.
bekas [8.4K]

Answer: A substance that produces hydroxide ions when placed in water is base.

Explanation:

Bases are the substance:

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  • BOH(aq)\rightarrow B^+(aq)+OH^-(aq)

Where as acid gives positively charged hydronium ion(H^+) in aqueous solution.

7 0
3 years ago
Read 2 more answers
A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0
3 years ago
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Effectus [21]

Milk

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6 0
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