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3 years ago

I NEED AN ANSWER TO THESE QUESTIONS ASAP!! I GOT A TEST TOMORROW ON THIS!!!

1 answer:

5 0

18.The octet rule tells us that in every chemical reactions, elements will either gain or lose electrons to attain the noble gas electron configuration. This stable electron configuration is known as the octet configuration since it is composed of 8 valence. Oxygen’s electron configuration is 1s2 2s2 2p4. So when oxygen reacts with other elements to form compounds, it completes the octet configuration by taking 2 electrons from the element it reacts with

19. Actually pure metals are made up not of metal atoms but rather of closely packed cations (positively charge particles). These cations are then surrounded by a pack of mobile valence electrons which drift from one part of the metal to another. This is called metallic bond.

20. This is the energy which is needed to break a single bond. When the dissociation energy is large, this means that the compound is more stable. Since carbon to carbon bonds have high dissociation energy, therefore they are not very reactive.

21. Network solids are type of solids in which the atoms are covalently bonded to one another, so they are very stable. It takes higher temperature to melt them because breaking these covalent bonds required greater energy. Some examples are:
- Diamond
-Silicon Carbide

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Please help with that problem Important İ have quiz tomorrow

iogann1982 [59]

Answer:

Final pressure of the gas remaining in the first container is 3.5 atm

Explanation:

Since it is given that temperature is constant , we can apply -

PV=constant , where

P = Pressure of the gas

V = Volume of the container in which the gas is contained

Initially,

For container 1 -

$P_{i}$ = 4 atm

$V_{1}$ = 6 L

Finally,

For container 2 -

$P_{f,2}$ = 3 atm

$V_{2}$ = 1 L

For container 1 -

$P_{f,1}$ = ?

$V_{1}$ = 6 L

∴ $P_{i}$ $V_{1}$ = $P_{f,1}$ $V_{1}$ + $P_{f,2}$ $V_{2}$

∴ 4×6 = ( $P_{f,1}$ ×6) +(3×1)

∴ $P_{f,1}$ = 3.5 atm

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3 years ago

Diffusion of nonpolar molecules would not be affected by

wariber [46]

+molecule size +steepness of the concentration gradient +temperature + steepness of the electric gradient +steepness

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3 years ago

Calculate the pH of a buffer prepared by mixing 20.0 mL of 0.10 M acetic acid and 55.0 mL of 0.10 M sodium acetate

MatroZZZ [7]

Answer:

Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.

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3 years ago

What is solution in chemistry

nika2105 [10]

a solution in chemistry is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.

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3 years ago

When the equation Eg(s) + DnGO4(aq) ---> Dn(s) + Eg2(GO4)3(aq) is balanced, what is the coefficient in

Artemon [7]

B) 2

You would first balance GO4 by adding a coefficient of 3 in front of DnGO4 in the reactants. Then you’d balance the 3DnGO4 by adding a coefficient of 3 in front of Dn in the products. Finally you’d balance Eg by adding a coefficient of 2 in front of Eg to balance with the Eg2(GO4)3 in the products.

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3 years ago