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3 years ago

Given the following list of atomic and ionic species, find the appropriate match for questions 2-4. Please Explain.

Chemistry

1 answer:

OleMash [197]3 years ago

4 0

Answer:

2. (C) K⁺; 3. (E) Hg⁺; 4. Hg⁺

Explanation:

We must first write the electron configurations of the different species.

(A) Fe²⁺

Fe: [Ar]4s²3d⁶

Fe²⁺: [Ar]3d⁶

When removing electrons from a transition metal ion, you remove the s electrons first.

(B) Cl

Cl: [Ne]3s²3p⁵

(C) K⁺

K: [Ar]4s

K⁺: [Ar]

(D) Cs

Cs: [Xe]6s

(E) Hg⁺

Hg: [Xe]6s²4f¹⁴5d¹⁰

Hg⁺: [Xe]6s4f¹⁴5d¹⁰

2. K⁺ has a noble gas configuration

3. Hg⁺ has electrons in f orbitals.

4. The electron configuration of Au is [Xe]6s4f¹⁴5d¹⁰, not [Xe]6s²4f¹⁴5d⁹, because a filled d subshell is more stable than a filled s subshell.

Thus, Hg⁺ is isoelectronic with Au.

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Which of Newton's Laws of motion requires you to have brakes on your bicycle?

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<span>Which of Newton's Laws of motion requires you to have brakes on your bicycle? The answer is the = 1st law</span>

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2 years ago

Cyclohexane, a commonly used organic solvent, is 85.6% c and 14.4% h by mass with a molar mass of 84.2 g/mol. what is its molecu

puteri [66]

The molecular formula of organic solvent is <em>C6H12</em>

<h2>calculation</h2><h3>find the empirical formula first as in step 1 and 2</h3>

Step 1: f<em>ind the moles of C and H</em>

moles = % composition/molar mass

from periodic table molar mass of C= 12 g/mol while that of H= 1 g/mol
moles is C is therefore = 85.6/12= 7. 13 moles
moles of H= 14.4/1 - 14.4 moles

Step 2: <em>calculate the mole fraction by dividing each mole by smallest number of mole(7.13)</em>

that is C= 7.13/7.13 = 1

H= 14.4/7.13 =2

the empirical formula is therefore = CH2

<h2>Then calculate the molecular formula from empirical formula</h2>

step 3: divide the grams molar mass by empirical formula mass

empirical formula mass = 12+(1 x2) = 14 g/mol

= 84.2/ 14 = 6

step 4: multiply each of the subscript within the empirical formula with the value gotten in step 3

that is [CH2]6 = C6H12 therefore the molecular formula = <u>C6H12</u>

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2 years ago

In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.

trasher [3.6K]

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

$V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}$

$10=\frac{4}{3}*pi*{r_{L}}^{3}$

${r_{L}}^{3}=\frac{7.5}{pi}$

$r_{L}=1.337cm$

Step 2: Determine surface area of large sphere

$A_{L}=4*pi*{r_{L}}^{2}$

$A_{L}=4*pi*1.337^{2}$

$A_{L}=22.463 cm^{2}$

Step 3: Determine radius of small sphere

$V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}$

$1.25=\frac{4}{3}*pi*{r_{s}}^{3}$

${r_{s}}^{3}=\frac{0.9375}{pi}$

$r_{s}=0.668cm$

Step 4: Determine surface area of small sphere

$A_{s}=4*pi*{r_{s}}^{2}$

$A_{s}=4*pi*0.668^{2}$

$A_{s}=5.607 cm^{2}$

Step 5: Determine total surface area of 8 small spheres

$A_{S}=8*A_{s}$

$A_{S}=8*5.607$

$A_{S}=44.856 cm^{2}$

$A_{L}=22.463 cm^{2}$ - Surface area of 1 large sphere

$A_{S}=44.856 cm^{2}$ - Surface area of 8 small spheres

Options:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm

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2 years ago

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anastassius [24]

Since the nucleophile is the actual attacking molecule or molecule that starts the reaction and allows for further steps in the mechanism to occur, it is the limiting reagent, as based on the amount of the nucleophile you have, the reaction will tend to proceed until you run out. The excess would be the sodium hydroxide, it is union part of the solution.

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3 years ago

Melting gold with other metals makes 10 karat gold. Which term best describes 10 karat gold

katrin2010 [14]

The 10 karat of gold formed by melting gold with other metals is described as homogeneous mixture.

A pure substance is a substance that is made up of only one kind of particle. This type of substance consist of only one element.

A heterogeneous substance is a type of substance that is made up two or more substances existing in different phases.

A homogeneous substance is a type of substance that is made up of two or more substances evenly mixed. This type of substance exist only in one phase.

Thus, the 10 karat of gold formed by melting gold with other metals is described as homogeneous mixture.

Learn more about homogeneous mixture here: brainly.com/question/10677519

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1 year ago

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