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sashaice [31]
3 years ago
8

what order does covalent bond, dipole-dipole, hydrogen bond, ionic bond, metalic bond, and london dispersion bond for strength?​

Chemistry
1 answer:
jarptica [38.1K]3 years ago
7 0

Answer:The major types of solids are ionic, molecular, covalent, and metallic. ... (network) , or metallic, where the general order of increasing strength of interactions. ... In ionic and molecular solids, there are no chemical bonds between the ... by dipole –dipole interactions, London dispersion forces, or hydrogen ...

Explanation:

Hope this will help you

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Air has a mass of 1.2g and a volume of 4,555ml. What is the density
coldgirl [10]

Answer:

<h2>Density = 0.00026 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>Density(\rho) =  \frac{mass}{volume}</h3>

From the question

mass of air = 1.2 g

volume = 4,555 mL

Substitute the values into the above formula and solve for the density

That's

<h3>Density =  \frac{1.2}{4555}</h3>

= 0.0002634

We have the final answer as

<h3>Density = 0.00026 g/mL</h3>

Hope this helps you

3 0
3 years ago
What is the atomic mass of copper?
kipiarov [429]

Answer:

63.546 u

Explanation:

4 0
3 years ago
A reaction is exothermic when Group of answer choices weak bonds break and strong bonds form. strong bonds break and weak bonds
Leona [35]

Answer:

weak bonds break and strong bonds form

Explanation:

An exothermic reaction is a chemical reaction in which heat energy is evolved during the reaction process.

Bond formation requires energy while bond breakage releases energy. More energy is needed for the formation of weak bonds as compared to strong bonds.

<em>Hence, when weak bonds break, they release more energy than needed to make a corresponding strong bond leading to the release of the remaining energy as heat.</em>

6 0
3 years ago
Assume that the pressure inside of the 1 liter balloon is 200,000 Pa, what is the explosive energy in the inside the balloon? Ex
AysviL [449]

Answer:

200 Joules is the explosive energy in the inside the balloon. And that is  9.523\times 10^{-5}1 lb of TNT.

Explanation:

Pressure=\frac{force}{Area}=\frac{Force\times distance}{Area \times distance}=\frac{Energy}{Volume}

Volume of the balloon = V = 1 L = 0.001 m^3

Pressure inside the balloon ,P= 200,000 Pa =200,000 N/m^2

Explosive energy in the inside the balloon be E.

E = Pressure × Volume

E=200,000 N/m^2\times 0.001 m^3=200 Joules

1 lb of TNT = 2.1\times 10^6 J

200 Joules = 200\times \frac{1}{2.1\times 10^6 }1 lb of TNT

= 9.523\times 10^{-5}1 lb of TNT

5 0
3 years ago
For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibr
lys-0071 [83]

Answer:

a. No change.    

b. The equilibrium will shift to the right.

c. No change

d. No change

e.  The equilibrium will shift to the left

f.  The equilibrium will shift to the right      

Explanation:

We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining   Keq  again.

The equilibrium constant  for  A(s)⇌B(g)+C(g)  

Keq = Kp = pB x pC

where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )

We also use  Q which has the same form as Kp but denotes the system is not at equilibrium:

Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.

a.  double the concentrations of Q which has the same form as Kp but : products and then double the container volume

Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.

Initially the system will decrease the partial pressures of B and C by a half:

Q = pB´x pC´     ( where pB´and pC´are the changed pressures )

Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp  ⇒ Kp = Q/4

But then when we double the volume ,the sistem will react to  double the pressures of A and B. Therefore there is no change.

b.  double the container volume

From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change  reduced the pressures by a half :

Q =  pB´x pC´  = (  1/2 pB ) x ( 1/2 pC )  =  1/4 pB x pC  = 1/4 Kp

c. add more A

There is no change in the partial pressures of B and C since the solid A does not influence the value of kp

d. doubling the  concentration of B and halve the concentration of C

Doubling the concentrantion doubles  the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:

Q =  pB´x pC´  = ( 2 pB ) x ( 1/2 pC ) = K

e.  double the concentrations of both products

We learned that doubling the concentration doubles the pressure so:

Q =  pB´x pC´   = ( 2 pB ) x ( 2 pC ) = 4 Kp

Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.

f.  double the concentrations of both products and then quadruple the container volume

We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:

Q =  pB´x pC´   = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16  Kp = 1/4 Kp

So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.

If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.

3 0
3 years ago
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