All categories

3 years ago

Metallic bonding is possible because the electrons

1 answer:

4 0

It is possible because the electrons in the energy level furthest from the nucleus delocalise, this leaves behind positively charged metal ions which are attracted to the negatively charged electrons that now move around the lattice freely

You might be interested in

Which component of the galaxy is the arrow pointing to? a galaxy with the arrow pointing to the space between the fixed points o

VARVARA [1.3K]

A would be correct

srry u had to wait long

8 0

2 years ago

Read 2 more answers

What direction do the plates have to move in order for an earthquake to occur?

Bess [88]

they would be pushing together

4 0

3 years ago

Read 2 more answers

Which two notations represent isotopes of the same element

Mumz [18]

The two notations that represent isotopes of the same element is the one that represented in option 1
The lower number is the number of protons while the upper number is the atomic weight

hope this helps

6 0

3 years ago

Read 2 more answers

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

Which graphic below shows the correct orientation of each water molecule when it is near the cl- ion in the aqueous solution?

andrew-mc [135]

The question is missing the graphics required to answer which I have attached as an image.

There are four different representations of the orientation of water molecules around chloride anion. Let's first analyze the water molecule.

We have H-O-H as the structure of water. The oxygen atom is more electronegative than the hydrogen atoms, which results in a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom.

The chloride anion is a negative charge. Therefore, the water molecules should orient themselves with the hydrogen atoms facing the chlorine atom as the partial positive charge on the hydrogen atoms will be attracted to the negative charge of the chlorine atom.

The correct representation is shown in graph 3 which shows all hydrogen atoms facing the chlorine anion.

8 0

3 years ago