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sineoko [7]
3 years ago
6

If the two atoms participating in a bond have an electron negativity difference of 1.9 , what type of of bond would they be part

icipating in ?
Chemistry
1 answer:
brilliants [131]3 years ago
7 0

Answer:

Polar covalent bond.

Explanation:

When the bond is formed between the atoms by sharing the electrons the bond thus have covalent character.   The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.  When the electronegativity difference is less than 0.4 the bond is non polar covalent.

When bonded atoms have greater electronegativity difference i.e 2 or greater than two the bond is ionic because electron is transfer from low electronegative atom to highest electronegative atom.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.

In case of H₂, Cl₂, Br₂ the bond has very high covalent character because of zero electronegativity difference.

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Anna11 [10]

Answer:it’s C

Explanation:

A distant luminous object travels rapidly away from an observer.

4 0
2 years ago
Which best describes a similarity between power plants that use water as an energy source and those that use wind as an energy s
elixir [45]

Answer:

Both use kinetic energy to produce electricity.

6 0
3 years ago
What is the pH of a 2.10 x 10⁻⁶ M solution of HCl?<br>​
Delvig [45]

Answer:

pH = 5.7

Explanation:

pH = -log[H^+]  

For HCl pH = -log[HCl] = - log [2.10 x 10⁻⁶ ] = 5.7

5 0
3 years ago
A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

4 0
3 years ago
A standard solution is prepared for the analysis of fluoxymesterone (c20h29fo3; 336 g/mol), an anabolic steroid. a stock solutio
-Dominant- [34]

<em>Answer:</em>

  • The concentration of new solution will be 1×10∧-7 M.

<em>Solution:</em>

<em>Data Given </em>

       given mass of fluoxymesterone =16.8mg = 0.0168 g

       molar mass  of fluoxymesterone = 336g/mol

       vol. of fluoxymesterone = 500.0 ml = 0.500 L

      Stock Molarity of  fluoxymesterone = (0.0168/336)÷0.500 = 1×10∧-4 M

So applying dilution formula

                   Stock Solution :  New Solution

                                 M1.V1 = M2.V2

       ( 1×10∧-4 M) × (1×10∧-6 L) = M2 × 0.001 L

     [( 1×10∧-4) × (1×10∧-6)]÷[0.001] = M2

     1 × 10∧-7 = M2

<em>Result:</em>

  • The concentration of new solution M2 will be  1 × 10∧-7
3 0
3 years ago
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