Ask question

Ask question

All categories

2 years ago

14

The attractive forces between sodium and chloride ions are called:

1 answer:

Vedmedyk [2.9K]2 years ago

3 0

Dipole-dipole forces are probably the simplest to understand. You probably already know that in an ionic solid like NaCl, the solid is held together by Coulomb attractions between the oppositely-charges ions. The Na+ and Cl- ions alternate so the Coulomb forces are attractive

You might be interested in

Pls pls I need help to solve this questions

cupoosta [38]

1) • Solubility is the ability of a substance to be dissolved especially in water
• Unit; mg/L (milligrams per liter)

2) A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent.

3) To detoxify any acid that was previously used in it

4) CAN’T SEE THE MAIN QUESTIONS

5) CAN’T SEE THE MAIN QUESTIONS

6 0

3 years ago

In transduction, the cochlea is part of this structure:and the retina is part of this structure:

KengaRu [80]

Answer:

loooooooooooooooooooooooooolzzzz

Explanation:

5 0

2 years ago

Read 2 more answers

How is a water wave similar to a pendulum?

Flura [38]

Answer:

side to side?

Explanation:

3 0

2 years ago

Yall i think if someone can solve ths problem would save my life. its really URGENT

andre [41]

Cutxitxrxuyrxxydxdyxt

6 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

2 years ago