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Y_Kistochka [10]
3 years ago
8

Compare a sodium ion and neutral sodium atom in size. what causes the size difference

Chemistry
1 answer:
n200080 [17]3 years ago
4 0
A cation is smaller than its parent atom.
Therefore, here a sodium ion is smaller than the neutral sodium atom.

Reason: Since the sodium atom loses an electron to form an electropositive ion(cation), its nuclear charge increases(the number of protons exceeds the number of electrons) and the orbital electrons are pulled with a greater energy.
Thus,the sodium ion shrinks in size.
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Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f
Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
Which of these statement(s) is/are true about a balloon filled with 1.00 mol N2 (g) at STP?
lara [203]

Answer:

I. The balloon has a volume of 22.4L

III. The balloon contains 6.022x10^23 molecules.

Explanation:

At stp, it has been proven that 1mole of a gas occupy 22.4L.

Therefore, option (i) is correct.

The molar mass N2 = 14.01 x 2 = 28.02g/mol

Number of mole of N2 = 1 mole

Mass of N2 =..?

Mass = mole x molar Mass

Mass of N2 = 1 x 28.02 = 28.02g.

The mass content of the balloon is 28.02g, therefore, option (ii) is wrong.

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules. This implies that 1 mole of N2 also contains 6.02x10^23 molecules

Therefore, option (iii) is correct.

The correct options to the question are:

Option i and option iii

7 0
3 years ago
Read 2 more answers
NaCl and HCl both have chlorine but are different types of bonds.
Tema [17]

Answer:

In HCl ,bond between hydrogen and chlorine is covalent. But due to difference in electronegativity of hydrogen and chlorine,shared electron pair gets attracted towards chlorine. Due to this polar covalent

3 0
3 years ago
Draw the structures of the two monomers that react together to form this polyurethane polymer
matrenka [14]

The monomers that react together to form polyurethane polymer are diisocyanate and diol.

<h3>What are polymers?</h3>
  • Polymers are a class of natural or synthetic substances
  • It is made up of small units called monomers arranged in a repeated pattern to form large molecules called macromolecules
  • Cellulose and resins are examples of natural polymers
  • Polyethylene and polychloroprene are examples of synthetic polymers.

Polyurethane polymer is made up of monomers diisocyanate and diol. It is mostly used in home furnishing in flexible form. The structures of monomers are as follows:

Learn more about polymers:

brainly.com/question/17638582

#SPJ4

7 0
1 year ago
You mix 265.0 mL of 1.20 M lead(II) nitrate with 293 mL of 1.55 M potassium iodide. The lead(II) iodide is insoluble. What amoun
slava [35]

Answer:

105 grams PbI₂

Explanation:

Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)

moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole

moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant

moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂

grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)

7 0
3 years ago
Read 2 more answers
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