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3 years ago

Which atom attracts electrons most strongly? br F Rb

Chemistry

2 answers:

vovikov84 [41]3 years ago

8 0

Flourine because it has less energy levels hence strong nuclear force of attraction

Brut [27]3 years ago

3 0

Answer:

fluorine atoms!!! can you please mark brainliest?

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4.a) Draw two Lewis structures for HCP; one where C is central and one where P is central: b) Calculate formal charge for each a

love history [14]

One Hydrogen atom (H) and one Oxygen atom (O) surround the central Carbon atom (C) in the HCP Lewis structure (O). Carbon (C) and Phosphorus (P) have a triple bond, and Carbon (C) and Hydrogen (H) have a single bond.

<h3>How can you choose the ideal format for a formal charge?</h3>

The Lewis structure with the negative formal charges on the most electronegative atoms is the one to choose from when faced with a choice between numerous Lewis structures with similar formal charge distributions.

<h3>How do you determine the preferred resonance structure?</h3>

The resonance forms with the fewest non-zero formal charge atoms are selected. Resonance develops atoms that have a negative formal charge or are the most electronegative are preferred.

To know more about Lewis structure visit:-

brainly.com/question/20300458

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8 0

10 months ago

3. What are the products of a neutralization reaction?

Nadusha1986 [10]

Answer:

Neutralization reactions occur when two reactants, an acid and a base, combine to form the products salt and water.

5 0

3 years ago

Read 2 more answers

A metal ion Mⁿ⁺ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 2.961 x 10¹⁶ Hz

Whitepunk [10]

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion ( $Li^+$ )

Explanation:

First of all

v=f*λ

In our case v=c

c=f*λ

λ=c/f

where:

c is the speed of light

f is the frequency

$\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m$

Using Rydberg's Formula:

$\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where:

R is Rydberg constant= $1.097*10^7$

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

$\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99$

z≅3

Atomic number is 3, So ion is Lithium ion ( $Li^+$ )

3 0

3 years ago

g Suppose 0.0350 g M g is reacted with 10.00 mL of 6 M H C l to produce aqueous magnesium chloride and hydrogen gas. M g ( s ) +

iren2701 [21]

Answer:

Mg will be the limiting reagent.

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Mg: 1 mole
HCl: 2 moles
MgCl₂: 1 mole
H₂: 1 mole

Being the molar mass of each compound:

Mg: 24.3 g/mole
HCl: 36.45 g/mole
MgCl₂: 95.2 g/mole
H₂: 2 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Mg: 1 mole* 24.3 g/mole= 24.3 g
HCl: 2 moles* 36.45 g/mole= 72.9 g
MgCl₂: 1 mole* 95.2 g/mole= 95.2 g
H₂: 1 mole* 2 g/mole= 2 g

0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.

Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:

$Molarity=\frac{number of moles of solute}{volume}$

in units $\frac{moles}{liter}$

then, the number of moles of HCl that react is:

$6 M=\frac{number of moles of HCl}{0.01 L}$

number of moles of HCl= 6 M*0.01 L

number of moles of HCl= 0.06 moles

Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?

$mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}$

mass of Mg= 0.729 grams

But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, <u><em>Mg will be the limiting reagent.</em></u>

7 0

3 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Delicious77 [7]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>