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3 years ago

4. How many electrons are in the electron shell closest to the nucleus?

Chemistry

2 answers:

Roman55 [17]3 years ago

5 0

Answer:

Explanation:

there are 2 electrons close to the nucleus

Serhud [2]3 years ago

5 0

Answer:

Explanation:

2*n^2, where n is the electron shell.

If we plug 1 in, we will get 2*1, which is 2.

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The radius of a uranium atom is 149 pm. How many uranium atoms would have to be laid side by side to span a distance of 4.96 mm?

nikitadnepr [17]

Answer:

4960000000 pm

Explanation:

4.96*1000000000= 4960000000

4 0

3 years ago

Read 2 more answers

What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr

Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in $BeCl_2$ is, $sp$

(2) The hybridization of central atom nitrogen in $NO_2$ is, $sp^2$

(3) The hybridization of central atom carbon in $CCl_4$ is, $sp^3$

(4) The hybridization of central atom xenon in $XeF_4$ is, $sp^3d^2$

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, $BeCl_2$

$\text{Number of electrons}=\frac{1}{2}\times [2+2]=2$

The number of electron pair are 2 that means the hybridization will be $sp$ and the electronic geometry of the molecule will be linear.

(2) The given molecule is, $NO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, $sp^2$ .

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, $sp^2$ .

(3) The given molecule is, $CCl_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, $XeF_4$

$\text{Number of electrons}=\frac{1}{2}\times [8+4]=6$

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

3 0

3 years ago

Calculate the destiny of the object that has a mass of 453 g and occupies the volume of 225 cm3

Elan Coil [88]

density = 2.013g/cm3

Explanation:

density = mass/volume

density = 453g/225cm3

density = 2.013g/cm3

7 0

3 years ago

This is a PERCENTAGE COMPOSITION by mass problem.

Norma-Jean [14]

Answer:

I think it will be d 12% it will be true

4 0

3 years ago

Read 2 more answers

How many moles are in 15.2 grams of Calcium (Ca)?

nalin [4]

Answer:

There are 0.379 moles of Ca in the given sample.

Explanation:

Hey there!

We're given that we have 15.2 grams of Ca (Calcium). We need to convert grams to moles.

There are many different conversion types for moles. A few examples are:

grams to moles
moles to grams
grams to particles
moles to liters

There are really endless possibilities for conversion factors. Since we want to find the number of moles in a certain sample, we first need to define what a mole is.

Mole - a unit of measurement in chemistry

A mole is one unit of a certain item. For instance, one cup would be equivalent to one mole. Moles are simply a hypothetical unit in chemistry that act as a placeholder.

Now, how do we find the number of moles?

We know that a mole of something is one of that substance. In this case, we're referring to one atom of Calcium.

Therefore, we need to reference a periodic table to understand what the mass of one atom of calcium is equivalent to. I'll be using the official AP Chemistry periodic table as issued by the College Board.

When we reference the periodic table, we see the mass of one atom of calcium is equivalent to 40.08 amu (atomic mass units).

We learn in chemistry that we can use the terms amu and grams interchangeably, so we have learned that one mole of calcium is equivalent to 40.08 grams.

How do we work this out?

Now, we need to find out how many moles are in 15.2 grams of calcium. We can use dimensional analysis to work this out.

Our starting unit will be 15.2 grams of Ca.

$\text{15.2 grams Ca}$

We want to convert this into moles, so we need to cancel out our grams value. Additionally, since we know that we have 40.08 grams of Ca in one mole, we can set these up as two ratios:

$\displaystyle \frac{\text{40.08 grams Ca}}{\text{1 mole Ca}}$
$\displaystyle \frac{\text{1 mole Ca}}{\text{40.08 grams Ca}}$

These ratios can be used to express the relationship between the moles of calcium and the mass of one calcium atom. However, we need our value of grams to cancel out, so since we will be using cross multiplication in our dimensional analysis, we need to use the ratio with grams on the bottom.

Now that we have selected our ratio, we can set up a multiplication problem of our two ratios:

$\displaystyle \text{15.2 grams Ca} \times \frac{\text{1 mole Ca}}{\text{40.08 grams Ca}}$

We assume that we have a denominator of 1 below 15.2 grams Ca.

Now, we perform our multiplication.

$\displaystyle \frac{\text{15.2 grams Ca} \ \times \ \text{1 mole Ca}}{\text{40.08 grams Ca}}$

We see that our remaining unit is going to be mole(s) Ca, so we can go ahead and drop our grams Ca unit.

$\displaystyle \frac{\text{15.2} \ \times \ \text{1 mole Ca}}{\text{40.08}}$

Now, we need to simplify a little bit.

$\displaystyle \frac{\text{15.2}}{\text{40.08}} = \text{? moles Ca}$

Finally, let's divide our fraction and evaluate our equation to find the value of moles Ca.

$\displaystyle \frac{15.2}{40.08} = \text{0.379241517 moles Ca}$

Are we done yet?

Not just yet! To complete our work, we need to find the number of significant figures we can use. We are given two values in which we can use to judge how many significant figures we are limited to.

15.2 -- Three significant figures
40.08 -- Four significant figures

In chemistry, we always use the least number of significant figures for precision calculations, so we need to round our answer to three significant figures.

$0.379241517 \approx 0.379$

Therefore, there are approximately 0.379 moles of Ca in the given sample.

7 0

3 years ago