Which describes any compound that has at least one element from group 17

lana66690 [7]

Answer:

A. atomic number

Explanation:

the mass number alone cannot uniquely identify an element. But the number of protons in the nucleus identifies it uniquely. In addition, if the atom is neutral, then the atomic number also equals the number of electrons in its orbitals.

3 0

3 years ago

What is the specific name for the electrons involved in bonding?

Mariana [72]

Answer:

valence electrons are the ones

4 0

3 years ago

If an atom of sodium has 11 protons, what is its approximate atomic mass?

bagirrra123 [75]

23. The atomic mass is calculated by adding the number of protons and neutrons. However, you cannot guess the number of neutrons. Refer to the periodic table. The atomic mass/mass no of Sodium is 23.

7 0

3 years ago

Determine the concentration of nh3(aq) that is required to dissolve 743 mg of agcl(s) in 100.0 ml of solution. the ksp of agcl i

AnnyKZ [126]

Answer:

1.1 M

Explanation:

The dissociation of $AgCl_{(s)}$ is as follows:

$AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$

Given Value for $K_{sp} = 1.77*10^{-10}$

The equation for the reaction for the formation of complex ion $Ag(NH_3)^+_2$ is :

$Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}$

The value of $K_f = 1.6*10^7$

If we combine both equation and find the overall equilibrium constant will be:

$AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$

$Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}$

$AgCl_{(s)}+2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}$

$K = (1.77*10^{-10})(1.6*10^7)$

$K = 0.00283$

If $[NH_3]$ = x M

The solubility of $AgCl_{(s)}$ in the $NH_3$ solution will be:

$x = 743*10^{-3}g * \frac{mol AgCl}{143.32g}*\frac{1}{0.1000L}$

$x = 0.0518 M$

Constructing an ICE Table; we have :

$AgCl_{(s)} + 2 NH_3{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}$

Initial (M) x 0 0

Change (M) -2 (0.0518) + 0.0518 + 0.0518

Equilibrium (M) x - 0.1156 0.0518 0.0518

Equilibrium constant;

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

$0.00283 = \frac{(0.0518)^2}{(x-0.1156)^2}$

$0.00283 = (\frac{0.0518}{x-0.1156})^2$

$x = 0.1156 + \sqrt{\frac{0.0518^2}{0.00283} }$

x = [NH₃] = 1.089 M

[NH₃] ≅ 1.1 M

8 0

3 years ago

How many electrons are needed to completely fill the second level energy? The third? The fourth?

irinina [24]

The maximum number of electrons that an energy level can hold is determined from the formula 2n^2 equals the total number, where n is the energy level. The first energy level holds 2*1^2=2 electrons while the second holds 2*2^2= 8 electrons.

3 0

4 years ago