A school bus drives 35 mph down the street and slows as it approaches the stop sign

Consider the following electron configurations to answer the questions that follow: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (ii

Ahat [919]

Option (i) would have the highest 2nd Ionization Energy.

Option (i) is Sodium.

Can be Written as 2, 8 , 1

For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.

Now After Removing that Electron...

Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.

Neon has a very stable Octet.

It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.

So

Option (i) [Sodium] has the highest 2nd Ionization Energy

6 0

3 years ago

Using the periodic table, determine the ion charges of the following families of elements if valence electrons were removed or a

kogti [31]

Group I=1+

Group VI=-2

Group III=3+

6 0

3 years ago

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The mole fraction of a non-electrolyte (MM 40.0 g/mol) in a saturated aqueous solution is 0.310. What is the molality of the sol

jeka57 [31]

<u>Answer:</u> The molality of non-electrolyte is 24.69 m

<u>Explanation:</u>

We are given:

Mole fraction of saturated aqueous solution = 0.310

This means that 0.310 moles of non-electrolyte is present.

Moles of water (solvent) = 1 - 0.310 = 0.690 moles

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of water = 0.690 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$0.690mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g$

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

Where,

$n_{solute}$ = Moles of solute (non-electrolyte) = 0.310 moles

$W_{solvent}$ = Mass of solvent (water) = 12.42 g

Putting values in above equation, we get:

$\text{Molality of non-electrolyte}=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of non-electrolyte}=24.96m$

Hence, the molality of non-electrolyte is 24.69 m

4 0

3 years ago

Which observation could you make based on stimuli to your photoreceptors?(1 point)

Montano1993 [528]

Answer:

last one which is d

Explanation:

thankskd

6 0

3 years ago

Identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid and the conjugate base for each of the following r

Vlad1618 [11]

Answer:

Acids → H₂CO₃ from equilibrium 1 and water, from equilibrium 2.

Bases → Water from equilibrium 1 and ammonia from equilibrium 2.

In 1st equilibrium, H₃O⁺ is the conjugate acid and HCO₃⁻ the conjugate base.

In 2nd equilibrium, NH₄⁺ is the conjugate acid, and OH⁻, the conjugate base.

Explanation:

By the Bronsted-Lowry you know that acids are the one that release protons and base are the ones that catch them.

For the first equilibrium:

H₂CO₃(aq) + H₂O(l) ⇄ H₃O⁺(aq) + HCO₃⁻(aq)

Carbonic acid is the acid → It donates the proton to water, so the water becomes the base. As H₂CO₃ is the acid, the bicarbonate is the conjugate base (it can accept the proton from water to become carbonic acid, again) and the hydronium is the conjugate acid (it would release the proton to become water).

For the second equilibrium:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺ (aq) + OH⁻(aq)

This is the opposite situation → Water relase the proton to ammonia, that's why water is the acid and NH₃, the base (it accepted to become ammonium). The NH₄⁺ is the conjugate acid (it can release the H⁺ to become ammonia) and the OH⁻ is the conjugate base (It can accept the proton to become water, again).

5 0

3 years ago