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4 years ago

Calcium has an atomic number of 20. How many electrons of a calcium atom are involved in chemical bonding?

Chemistry

1 answer:

NARA [144]4 years ago

3 0

The answer is 2 electrons.

The electron configuration of calcium is 2:8:8:2

Calcium has two electrons in its outermost shell. These are its valence electrons and are the ones used in bonding with other elements. Valence electrons of an atom are those electrons that are in its outer energy shell or that are available for bonding.

Calcium is a metal. When metals react with non-metals, electrons are transferred from the metal atoms to the non-metal atoms forming ions. The resulting compound is known as an ionic compound.

For example, when calcium metal reacts with chlorine gas, calcium gives up its two valence electrons and Chlorine accepts them resulting in a new substance called calcium chloride in which the two elements have ended up forming ionic bonds.

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At 25 °c, only 0.0100 mol of the generic salt ab2 is soluble in 1.00 l of water. what is the ksp of the salt at 25 °c? ab2(s)↽−−

Assoli18 [71]

Where the solubility product Ksp is applied where salts don't fully dissolve in a solvent.

and when the reaction equation is:

AB2(s) ↔ A2+(Aq) + 2 B-(aq)

so we have the Ksp expression = [A2+][B-]^2

when we assume [A2+] = X = 0.01 M

and [B-] = 2X = 2*0.01 M = 0.02 M

So by substitution:

Ksp = 0.01 * (0.02)^2

= 4 x 10^-6

4 0

3 years ago

A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solut

leonid [27]

Answer:

x(t) = −39e

−0.03t + 40.

Explanation:

Let V (t) be the volume of solution (water and

nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid

measured in liters after t minutes, and let c(t) be the concentration (by volume) of

nitric acid in solution after t minutes.

The volume of solution V (t) doesn’t change over time since the inflow and outflow

of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is

c(t) = x(t)

V (t)

x(t)

200

We model this problem as

dt = I(t) − O(t),

where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,

both measured in liters of nitric acid per minute. The input rate is

I(t) = 6 Lsol.

1 min

20 Lnit.

100 Lsol.

120 Lnit.

100 min

= 1.2 Lnit./min.

The output rate is

O(t) = (6 Lsol./min)c(t) = 6 Lsol.

1 min

x(t) Lnit.

200 Lsol.

3x(t) Lnit.

100 min

= 0.03 x(t) Lnit./min.

The equation is then

dt = 1.2 − 0.03x,

dt + 0.03x = 1.2, (1)

which is a linear equation. The initial condition condition is found in the following

way:

c(0) = 0.5% = 5 Lnit.

1000 Lsol.

x(0) Lnit.

200 Lsol.

Thus x(0) = 1.

In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is

µ(t) = exp Z

P(t) dt

= exp

0.03 Z

= e

0.03t

The solution is

x(t) = 1

µ(t)

µ(t)Q(t) dt + C

= Ce−0.03t + 1.2e

−0.03t

0.03t

= Ce−0.03t +

1.2

0.03

−0.03t

0.03t

= Ce−0.03t +

1.2

0.03

= Ce−0.03t + 40.

The constant is found using x(t) = 1:

x(0) = Ce−0.03(0) + 40 = C + 40 = 1.

Thus C = −39, and the solution is

x(t) = −39e

−0.03t + 40.

3 0

3 years ago

HELP ASAP

sergejj [24]

D- A molecule

Explanation:

(Apex)

3 0

3 years ago

In the mass spectrum of the molecule phenol, C6H5OH, the approximate intensity of the peak at m/z 95, relative to the molecular

klasskru [66]

The correct answer to thi

6 0

4 years ago

determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth m

mamaluj [8]

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) $1s^22s^22p^63s^23p^5$

(b) $1s^22s^22p^63s^23p^63d^74s^2$

(d) $1s^22s^22p^63s^23p^63d^4s^1$

Answer :

(a) $1s^22s^22p^63s^23p^5$ → Halogen

(b) $1s^22s^22p^63s^23p^63d^74s^2$ → Transition metal

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$ → Transition metal

(d) $1s^22s^22p^63s^23p^63d^4s^1$ → Transition metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: $ns^2np^6$ where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: $ns^2np^5$ where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: $ns^1$ where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: $ns^2$ where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: $(n-1)d^{1-10}ns^{0-2}$ where n is the outermost shell.

(a) $1s^22s^22p^63s^23p^5$

The element having this electronic configuration belongs to the halogen family.

(b) $1s^22s^22p^63s^23p^63d^74s^2$

The element having this electronic configuration belongs to the transition family.

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

The element having this electronic configuration belongs to the transition family.

(d) $1s^22s^22p^63s^23p^63d^4s^1$

The element having this electronic configuration belongs to the transition family.

4 0

3 years ago