12 Significant Figures in Addition and

There are some exceptions to the trends of first and successive ionization energies. For each of the following pairs, explain wh

NNADVOKAT [17]

Answer:

(b) IE₂ of Ga > IE₂ of Ge

Explanation:

Electronic configuration of Ga is [Ar] 3d¹⁰4s²4p¹

Electronic configuration of Ge is [Ar] 3d¹⁰4s²4p²

After 1st ionisation , Ga becomes [Ar] 3d¹⁰4s² and becomes stable . Its

2 nd ionisation requires higher amount of ionisation energy. In case of Ge , there are 2 electrons in outermost orbital so it becomes stable after ionisation of 2 electrons.

4 0

3 years ago

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How many moles of mercury are equivalent to 3.46 x 1023 atoms?

olga55 [171]

Answer:

3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.

4 0

2 years ago

What are some examples of minerals that occur as elements instead of<br>compounds

katrin [286]

Answer: Aluminum, Antimony, Arsenic, Bismuth, Carbon, Cadmium, Chromium, Cobalt, etc.

3 0

3 years ago

Liquid water can be separated into hydrogen gas and oxygen gas through electrolysis. 1 mole of hydrogen gas and 0.5 moles of oxy

soldier1979 [14.2K]

The temperatures of the gases will not be equal as oxygen gas will have a higher temperature than hydrogen gas because it has fewer moles overall.

<h3>Briefing :</h3>

The mechanical behavior of ideal gases is described by the ideal gas law. It has the ability to compute the volume of gases created or absorbed.

This equation is frequently used in chemical equations to convert between volumes and molar quantities.

According to the ideal gas law, there is a relationship between gas pressure, temperature, and volume.

PV = nRT

V is the same for both

So,

T is same for both.

When n increases, T decreases, so since n for hydrogen gas is 1 and n for oxygen gas is 0.5, it follows that oxygen gas will have a higher temperature than hydrogen gas because it has fewer moles overall.

To know more about ideal gases :

brainly.com/question/15962335

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7 0

1 year ago

A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of

pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = <em>0.20</em>

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I hope it helps!

3 0

3 years ago