All categories

1 year ago

State the type of bonding—ionic, covalent, or metallic—you would expect in each: (b) MgCl₂(s);

1 answer:

3 0

MgCl2 is an ionic compound because chemical bonds in the molecule are formed by the transfer of electrons among Mg and Cl atoms.

<h3>What is chemical bond ?</h3>

A chemical bond is a strong bond that can be formed between atoms, ions, or molecules to create chemical compounds. The bond may be created by the sharing of electrons in covalent bonds or by the electrostatic attraction of two oppositely charged ions, as in ionic bonds. Covalent, ionic, and metallic bindings are examples of "strong bonds" or "primary bonds," whereas dipole-dipole interactions, the London dispersion force, and hydrogen bonding are examples of "weak bonds" or "secondary bonds."

The positively charged protons in the nucleus and the negatively charged electrons in its orbit are attracted to one another by the basic electromagnetic force.

To learn more about chemical bonds from the given link:

brainly.com/question/819068

#SPJ4

You might be interested in

Copper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f

Kruka [31]

Answer:

53.18 gL⁻¹

Explanation:

Given that:

$Cu^{2+}_{(aq)}$ $+$ $2NH_{3(aq)}$ $------>$ $[Cu(NH_3)_2]^+_{(aq)}$ ------equation (1)

where;

Formation Constant $(k_f) =$ $6.3*10^{10}$

However, the Dissociation of $CuBr_{(s)$ yields:

$CuBr_{(s)}$ ⇄ $Cu^{+}_{(aq)}$ $+$ $Br^-_{(aq)}$ -------------- equation (2)

where;

the Solubility Constant $(k_{sp})$ $= 6.3 *10^{-9$

From equation (1);

$(k_f) =$ $\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}$ --------- equation (3)

From equation (2)

$(k_{sp})$ $= [Cu^+][Br^-]$ --------- equation (4)

In $NH_3$ , the net reaction for $CuBr_{(s)$ can be illustrated as:

$CuBr_{(s)$ $+$ $2NH_{3(aq)}$ ⇄ $[Cu(NH_3)_2]^+_{(aq)}$ $+ Br^-_{(aq)}$

The equilibrium constant (K) can be written as :

$K=$ $\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}$

If we multiply both the numerator and the denominator with $[Cu^+]$ ; we have:

$K=$ $\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}$

$K=$ $\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}$

$K = k_f *k_{sp}$

$K= (6.3*10^{10})*(6.3*10^{-9})$

$K= 3.97*10^2$

$K$ ≅ $4.0*10^2$

Now; we can re-write our equilibrium constant again as:

$K=$ $\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}$

$4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}$

$4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}$

$4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2$

By finding the square of both sides, we have

$\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2$

$2.0*10 = \frac{x}{(0.76-2x)}$

$20(0.76-2x) =x$

$15.2 -40x=x$

$15.2 = 40x +x$

$15.2 = 41x$

$x = \frac{15.2}{41}$

$x = 0.3707 M$

In gL⁻¹; the solubility of $CuBr_{(s)$ in 0.76 M $NH_3$ solution will be:

$= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}$

= 53.18 gL⁻¹

4 0

3 years ago

Which of the following pairs of substances can be broken down chemically?

Ket [755]

Methane and water because they are both compounds

5 0

2 years ago

When thermal energy is added to an object, what happens to the motion of the particles?

Novay_Z [31]

The atoms start vibrating faster

3 0

2 years ago

Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives. The first step in its synthesis is th

Hunter-Best [27]

Answer:

The answer is 0.36 kg/s NO

Explanation:

the chemical reaction of NH3 to NO is as follows:

4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)

We have the following data:

O2 Volume rate = 645 L/s

P = 0.88 atm

T = 195°C + 273 = 468 K

NO molecular weight = 30.01 g/mol

we calculate the moles found in 645 L of O2:

P*V = n*R*T

n = P*V/R*T

n= (0.88 atm * 645L/s)/((0.08205 L*atm/K*mol) * 468 K) = 14.78 moles of O2

With the reaction we can calculate the number of moles of NO and with its molecular weight we will have the rate of NO:

14.78 moles/s O2 * 4 molesNO/5 molesO2 * 30.01 g NO/1 molNO x 1 kgNO/1000 gNO = 0.36 kg/s NO

8 0

3 years ago

Kp for the following reaction is 0.16 at 25 degree C. 2 NOBr(g) 2 NO(g) Br_2(g) The enthalpy change for the reaction at standard

finlep [7]

Answer:

Explanation:

Given that:

$2 NOBr_{(g)} \iff 2 NO_{(g)} + Br_{2(g)}$

From above:

$K_p = 0.16 = \dfrac{(P_{NO})^2 (P_{Br})}{(P_{NOBr})^2}$

To predict the effect of the addition of Br₂(g);

The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr

The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.

5 0

2 years ago