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3 years ago

An atom of which of these elements most likely forms an anion? A. AI B. Br C. Ca D. Mg

2 answers:

8 0

Bromine because it has seven electrons in its outer shell. This means that it needs to gain an electron in order to have a full outershell (eight electrons). A negative ion is called an anion.

Contact [7]3 years ago

4 0

Answer: The correct answer is Option B.

Explanation:

Anions is the negatively charged ion which is formed when an element gains electrons.

Cations are the positively charges ions which is formed when an element looses electrons.

From the given options:

Option A: Aluminium (Al)

Aluminium is the 13th element of the periodic table having electronic configuration of $1s^22s^22p^63s^23p^1$

This element will loose 3 electrons and will form $Al^{3+}$ ion known as cation.

Option B: Bromine (Br)

Bromine is the 35th element of the periodic table having electronic configuration of $[Ar]3d^{10}4s^24p^5$

This element will gain 1 electron and will form $Br^-$ ion known as anion.

Option C: Calcium (Ca)

Calcium is the 20th element of the periodic table having electronic configuration of $[Ar]4s^2$

This element will loose 2 electrons and will form $Ca^{2+}$ ion known as cation.

Option D: Magnesium (Mg)

Magnesium is the 12th element of the periodic table having electronic configuration of $1s^22s^22p^63s^2$

This element will loose 2 electrons and will form $Mg^{2+}$ ion known as cation.

From the above information, the correct answer is Option B.

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3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

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3 years ago

Help Please! Thanks!

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Answer:

Explanation:

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