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Katyanochek1 [597]
3 years ago
5

Do all dipole-dipole forces have the same strength?

Chemistry
1 answer:
n200080 [17]3 years ago
6 0
Inter molecula force 
1- hydrogen bond 
2- ion bond 
3- dipole dipole 
4- dipole bond or van der wael 
but not all dipole dipole has the same strength 
You might be interested in
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
3 years ago
Is table salt, NaCl, a typical ionic compound? Explain.
romanna [79]
Yes because Na is a metal and Cl is a nonmental, so they criss cross charges, as Na is 2+ and Cl is 2-, 2+ indicates that Na gives AWAY electrons to Cl and the 2- on Cl indicates it receiving electrons. NaCl is table salt and forms a crystal lattice structure, which is why salt is solid at room temperature. 

6 0
3 years ago
A fossil of an organism that lived during only one time is found in only one kind of rock layer, or stratum. If two strata conta
Damm [24]
I think this is true because they’re in the same layer, strata which would mean they died around the same time. Hope this helps sorry if it’s wrong!
6 0
3 years ago
Read 2 more answers
5.00 g of carbon were adiabatically burned to CO2 in a 2.00 kg copper calorimeter which contained 2.50 kg of water. The temperat
lorasvet [3.4K]

Answer:

A. -163.96kJ

B. -158.34kJ

Explanation:

Heat of combustion is the heat released when an element with oxygen at stp.

Reaction for the combustion of carbon:

C(s) + O2(g) --> CO2(g)

Enthalpy heat of combustion, C (using Hess law) = -393.5 kJ/mol

m = 5g

Molecular weight = 12 g/mol

No of moles = mass/molecular weight

= 5/12

= 0.417mol

DH = -393.5 * 0.417

= -163.96 kJ

B. Heat absorbed by the calorimeter = heat evolve by combustion

= mCH2ODT + mCcDT

Where CH2O is the specific heat capacity of the water in the calorimeter

Cc is the specific heat capacity of copper calorimeter

= (2500*4.184*14.1) + (2000*0.385*14.1)

= -158343J

= -158.34kJ

3 0
3 years ago
Please I really need help...
IrinaK [193]

Answer:

1

Explanation:

3 0
3 years ago
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