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vekshin1
3 years ago
9

Given an element with seven valence electrons, how many chemical bonds will usually form?

Chemistry
2 answers:
aleksandrvk [35]3 years ago
5 0

Answer : The element is belongs to the group 17 having the seven valence electrons and they will usually form one chemical bonds to another element.

Explanation :

As per given in the question that the element has seven valence electrons that means the seven electrons are present in their outermost energy level or shell.

Form the given information we conclude that, the group number is, 17 (because there are 7 valence electrons in outermost shell).

The elements belongs to group 17 are fluorine, chlorine, bromine and iodine and they are the halogens.

The general electronic configuration of halogens are, ns^2np^5 that means they have seven valence electrons in their outermost shell and they need only one electrons to complete their octet. So, they will usually form one chemical bonds by accepting one electrons from the other elements.

Hence, the element is belongs to the group 17 having the seven valence electrons and they will usually form one chemical bonds to another element.

Anna35 [415]3 years ago
4 0

Answer:

a

Explanation:

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2. Changing the compound changes the absorbance behavior.

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Explanation:

When the beverage company adds dye to the drink, there should be standard quantity added to the drink so that the color of the drink remains constant. When too much dye is added to the drink, the color will get dark brown or black. When the color of drink get lighter than green this means dye is not added in required quantity.

7 0
3 years ago
The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions
mamaluj [8]

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a) 62.1 kJ/mol

b) 2.82 kJ/mol

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d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

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ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

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Ans: B . 1 mole of gas

Explanation:

...

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