Which is a binary ionic compound?

2 answers:

8 0

A binary ionic compound is composed of ions of two different elements. One of which is a metal, and the other a non-metal

sasho [114]3 years ago

6 0

Answer:

I couldn't find options online for your question. But I can explain a little bit about it so you can identify the answer for yourself.

A binary ionic compound is made up of ions that have two different elements.

Of these two ions, one must be a metal and the other a nonmetallic.

The metallic atom will lose electrons, becoming a positive ion (cation), and the non-metallic atom will gain electrons, transforming into a negative ion (anion). Between them there is a very strong electrostatic attraction that is what holds them together.

Here are some examples of these types of compounds:

Sodium oxide: Na2O

Calcium hydroxide: Ca (OH) 2

Aluminum hydride: AlH3

Hydrochloric acid: HCl

You might be interested in

Paul added a sample of bromic acid to water and found the pOH to be 10. What is the pH of the solution?

Ket [755]

Answer:

Explanation:

pH + pOH = 14

pH + 10 = 14

pH = 14 - 10 = 4

Therefore pH of the solution is 4 .

6 0

2 years ago

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$