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3 years ago

The bond between sodium (na) and chlorine (cl) to form halite (salt) is a(n) ________ bond.

1 answer:

3 0

When an ionic bond is formed between sodium (na) and chlorine (cl), the resulting molecule is called halite (salt). Ionic bonds involve the transfer of valence electrons. When this occurs between metals and non-metals, the metal loses an electron (becoming a cation) and the non-metal gains an electron (becoming an anion).

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Explain why food chains do not tend to exceed four links.

Softa [21]

Energy is "lost" at each trophic level when you go up the chain. Typically there are fewer organisms at higher trophic levels.

Hope this helps!

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3 years ago

Unknown element x is a metal that ionically bonds to sulfur.is the formula, x3s feasible? why or why not?a)it is feasible. the t

nasty-shy [4]

The correct answer is:
d) No, it is not feasible. three metallic ions cannot provide the exact number of electrons that one sulfur needs for the ionic bond.
Because Sulfur is divalent so it need to gain 2 electrons from metal so if we have 3 metals they can't provide only two electrons only.

8 0

3 years ago

Read 2 more answers

The equilibrium constant for the reaction

FinnZ [79.3K]

Answer: The concentrations of $Cl_2$ at equilibrium is 0.023 M

Explanation:

Moles of $Cl_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol$

Volume of solution = 1 L

Initial concentration of $Cl_2$ = $\frac{0.14mol}{1L}=0.14M$

The given balanced equilibrium reaction is,

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial conc. 0.14 M 0 M 0M

At eqm. conc. (0.14-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}$

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

$4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}$

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of $Cl_2$ at equilibrium is 0.023 M

7 0

3 years ago

Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th

artcher [175]

Answer:

The equilibrium constant is:

$k_c=0.0030M^{-2}$

Explanation:

The correct equation is:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

$k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}$

Substituting:

$k_c=\dfrac{(0.105M)^2}{(1.1M)\cdot (1.50M)^3}$

$k_c=0.0030M^{-2}$

6 0

3 years ago

16. As he developed his periodic table, Mendeleev grouped the elements lithium,

Aleonysh [2.5K]

Answer:

the reactivity and the valence electrons

Explanation:

the reactivity of the elements would have played a significant role in why such elements were grouped together. the number of valence electrons dictates how reactive an element is - the less valence electrons the more reactive it is. the column, group 1 in which these elements are put together in, show that each of the elements have 1 valence electrons and are therefore reactive.

you can go on to further explain what valence electrons are, explain what the group numbers are associated with the valence electrons and how valence electrons effect reactivity. further this, talk about how the three elements have the same number of valence electrons and therefore were grouped together

8 0

3 years ago