Which of the following is NOT a type of pseudoscience?

Represent the formation of cations for the following metal atoms using electron dot structures. (a) Al (b) Sr (c) Ba

vagabundo [1.1K]

the formation of cations by using electron dot structures are :

a) Al

.

Al . losing the three valence electrons makes the Al³⁺

.

b) Sr :

Sr : losing the two valence electrons makes Sr²⁺

c) Ba

: Ba , losing the two valence electrons makes it Ba²⁺

A Lewis electron dot diagram is a representation of the valence electrons of an atom that employments specks around the image of the element. The number of dots equals the number of valence electrons within the molecule. These dots are arranged to the right and left and over and underneath the symbol, with no more than two dots on a side. Cations are the positive ions shaped by the loss of one or more electrons. The foremost commonly shaped cations of the representative elements are those that include the loss of all of the valence electrons.

To know more about the lewis electron dot diagram refer to the link brainly.com/question/14191114?referrer=searchResults.

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3 0

11 months ago

Tin has ten stable isotopes. The heaviest, 124Sn, makes up 5.80% of naturally occuring tin atoms. How many atoms of 124Sn are pr

matrenka [14]

The average atomic mass of Sn is 118.71 g/mol

the percentage of heaviest Sn is 5.80%

the given mass of Sn is 82g

The total moles of Sn will be = mass / atomic mass = 82/118.71=0.691

Total atoms of Sn in 82g = $6.023X10^{23}X0.691=4.16X10^{23}$

the percentage of heaviest Sn is 5.80%

So the total atoms of $Sn^{124}$ = 5.80% X $4.16X10^{23}$

Total atoms of $Sn^{124}$ = $2.41X10^{22}$ atoms

the mass of $Sn^{124}$ will be = $\frac{2.41X10^{22}X124}{6.023X10^{23}}=4.96g$

5 0

3 years ago

How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -><br> 4A1 + 302

Evgesh-ka [11]

<h3>Answer:</h3>

$\displaystyle 40 \ mol \ Al$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 40 \ mol \ Al$

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

4 0

2 years ago

draw the organic product formed when 1−hexyne is treated with h2o, h2so4, and hgso4. click the draw structure button to launch t

V125BC [204]

The organic product formed when 1−hexyne is treated with H₂O, H₂SO₄, and HgSO₄ will be 2-hexanone (structure attached).

This reaction is an example of an oxymercuration reaction of the organic product 1−hexyne.

Oxymercuration is shown in three steps to the right. The nucleophilic double bond attacks the mercury ion, releasing an acetoxy group. The mercury ion's electron pair attacks carbon on the double bond, generating a positive-charged mercuronium ion. Mercury's dxz and 6s orbitals give electrons to the double bond's lowest unoccupied molecular orbitals.

In the second stage, the nucleophilic H₂O attacks the highly modified carbon, freeing its mercury-bonding electrons. Electrons neutralize mercury ions by collapsing. Water molecules have positive-charged oxygen.

In the third stage, the negatively charged acetoxy ion released in the first step attacks the hydrogen of the water group, generating the waste product HOAc. The two electrons in the oxygen-hydrogen link collapse into oxygen, neutralizing its charge and forming alcohol.

You can also learn about organic products from the following question:

brainly.com/question/13513481

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4 0

1 year ago

Consider the following reaction: 2{\rm{ N}}_2 {\rm{O(}}g)\; \rightarrow \;2{\rm{ N}}_2 (g)\; + \;{\rm{O}}_2 (g)

fredd [130]

Answer:

a. 5.9 × 10⁻³ M/s

b. 0.012 M/s

Explanation:

Let's consider the following reaction.

2 N₂O(g) → 2 N₂(g) + O₂(g)

a.

Time (t): 12.0 s

Δn(O₂): 1.7 × 10⁻² mol

Volume (V): 0.240 L

We can find the average rate of the reaction over this time interval using the following expression.

r = Δn(O₂) / V × t

r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s

r = 5.9 × 10⁻³ M/s

b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:

5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s

4 0

3 years ago