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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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Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

lions [1.4K]

Answer: The equilibrium concentration of $COF_2$ is 0.332 M

Explanation:

We are given:

Initial concentration of $COF_2$ = 2.00 M

The given chemical equation follows:

$2COF_2(g)\rightleftharpoons CO_2(g)+CF_4(g)$

Initial: 2.00

At eqllm: 2.00-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}$

We are given:

$K_c=6.30$

Putting values in above expression, we get:

$6.30=\frac{x\times x}{(2.00-2x)^2}\\\\x=0.834,1.25$

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible

So, equilibrium concentration of $COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M$

Hence, the equilibrium concentration of $COF_2$ is 0.332 M

4 0

3 years ago

Write a reflection about your learning in this unit. Your reflection should be at least 3 sentences. Use the following sentence

Vera_Pavlovna [14]

Answer:

This unit has encouraged a deeper understanding of the world and it's guiding principles. While it was initially challenging for me to determine if a change was physical or chemical, this unit provided me with the information necessary to determine the type. With this knowledge, I can now interrelate with other properties and believe that this new ability will assist in future units as well. Thank you!

Explanation:

5 0

2 years ago

300 moles of sodium nitrite are needed for a reaction. the solution is 0.450 m. how many ml are needed?

miv72 [106K]

Answer:

$\boxed{\text{667 000 mL}}$

Explanation:

$\text{Molar concentration} = \dfrac{\text{moles}}{\text{litres}}\\\\c = \dfrac{n}{v}$

Data:

c = 0.450 mol·L⁻¹

n = 300 mol

Calculation:

$\begin{array}{rcl}0.450 & = & \dfrac{300}{V}\\\\0.450V& = & 300\\\\V & = & \dfrac{300}{0.450}\\\\V & = & \text{667 L} =\textbf{667 000 mL}\end{array}\\\text{The volume of solution needed is }\boxed{\textbf{667 000 mL}}$