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nata0808 [166]
2 years ago
6

Do the elements in group 7 of the periodic table react with metals to form covalent compounds, ionic compounds, neither, or both

?
Chemistry
1 answer:
Vladimir [108]2 years ago
7 0

Answer:Group 1 elements have 1 valence electron, meaning they have 1 extra electron that can easily be donated to an atom in search of 1 more electron. When they give away that extra electron to form an ionic compound, they become more stable.

For example, Group 7A (Group 17) elements have 7 valence electrons, meaning they need 1 extra electron to be stable. Group 1 and 7A elements make splendid ionic compounds.

Explanation:

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What is a possible set of quantum numbers for the unpaired electron in the following electron configuration: [Ar] 3d10 4s2 4p5?
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Answer:E

Explanation:

The electron is expected to be in any of the 4p orbitals. For the 4p orbitals

n=4

l=1

ml=-1,0,1

ms=-1/2 or +1/2

All the ml and ms configurations are equiprobable hence the answer.

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4 years ago
How many rings/energy levels would the Bohe model for an element
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It would have THREE rings/energy levels. This is because of the amount of electrons each orbital can hold.
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3 years ago
Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

Now put all the given values in this expression, we get :

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

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One method of increasing the chances of particles colliding during a reaction is to...
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Answer:

3

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