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2 years ago

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride ? atoms of boron.

Chemistry

1 answer:

maks197457 [2]2 years ago

5 0

Answer:

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.

2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.

Explanation:

The molecular formula of boron trifluoride is $BF_3$ .

So, one mole of boron trifluoride has one mole of boron atoms.

1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.

The number of atoms in 2.20 moles of boron is:

One mole of boron has ---- $6.023x10^2^3$ atoms.

Then, 2.20 moles of boron has

- $=2.20 mol. x 6.023 x 10^2^3 atoms /1 mol\\=13.25x10^2^3 atoms$

2. Calculate the number of moles of BF3 in 5.35*1022 molecules.

$(5.35x10^2^2 molecules/6.023x10^2^3)x 1mol\\=0.0888mol$

One mole of boron trifluoride has three moles of fluorine atoms.

Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.

=0.266mol of fluorine atoms.

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Which one of the samples contains the most molecules?

Alex

Answer: D is right

Explanation: One mole contains 6.0225 ·10^23 molecules

Despite of the substance

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3 years ago

How many O atoms are in 4C₆H₁₂O₆

ipn [44]

Number of O atoms : 24

<h3>Further explanation</h3>

Given

C₆H₁₂O₆ compound

Required

Number of atoms

Solution

A molecular formula shows the number of atomic elements in compound.

The empirical formula is the smallest comparison of the atoms

Glucose-C₆H₁₂O₆ is composed of 3 elements, namely C, H, and O.

The number of atoms in a compound can usually be seen from the subscript number after the atom and the reaction coefficient shows the number of molecules

So number of O atoms :

= 4 x 6 = 24 atoms

6 0

3 years ago

The quantum numbers for the last electron placed in three elements are listed below. Which of these is(are) NOT correct? Er (4 3

UNO [17]

Answer:

The three elements Erbium, Thallium and Osmium have incorrect quantum numbers for the last electron placed.

Explanation:

The 4 quantum numbers are (n,l,ml,ms):

n (Principal quantum number): it is the number of the shell (level) where the electron is placed.
l (Angular momentum quantum number or Secondary): it represents the sublevel where the electron is placed. There are 4 subleves: s, p d and f so secondary quantum number can take the number 0 (s), 1 (p), 2 (d) or 3 (f) depending on which sublevel the electron is placed.
ml (Magnetic quantum number): it represents the spatial orientation of the electron in respect of the sublevel the electron is placed. For example: if the electron occupies the s sublevel the magnetic number will be 0, if the electron occupies the p sublevel the magnetic number could be -1,0,1, if the electron occupies the d sublevel the magnetic number could be -2,-1,0,1,2 and if the electron occupies the f sublevel the magnetic number could be -3,-2,-1,0,1,2,3. You can see this in the attachment related to the correct sublevel for the example.

ms (Spin quantum number): this number represents the possible rotation of the electron so it could be 1/2 (which is represented by an up arrow) or -1/2 (represented by an down arrow).

Let's analyze the last electron of each element. You can see the attachment for better understanding. The last electron it is represented with orange color.

- Erbium:

This element has 68 electrons so following the Moeller's Diagram to fill the the electronic configuration, we found that the last electron of Erbium it is in the 4th level (shell), in the f sublevel. As Erbium has 12 electrons in the f sublevel, it is necessary to follow the Hund's rule (electrons must be placed singly in every sublevel before place a parallel electron) to placed correctly all of them. Finally, the last electron of Erbium stays in the middle of the sublevel and it is represented by a down arrow so the correct quantum numbers in the Erbium element are (4,3,1,-1/2).

- Thallium:

This element has 81 electrons and following the Moeller's Diagram, we found that it last electron it is in the 6th level, in the p sublevel. As Thallium has 1 electron in the p sublevel, it is placed singly in the sublevel. So the last electron of Thallium it is represented by an up arrow so the correct quantum numbers in the Thallium element are (6,1,-1,1/2).

- Osmium:

Osmium has 76 electrons and following the steps that we did with we the other elements, we noticed that its last electron it is in the 5th level, in the d sublevel. Following the Hund's rule the last electron of Osmium has a magnetic quantum number of -2 and its spin quantum number is -1/2, so the quantum numbers in the Osmium element are (5,2,-2,-1/2).

Note:

- Remember that the s sublevel has place for 2 electrons, the p sublevel has place for 6 electrons, the d sublevel has place for 10 electrons and the f sublevel has place for 14 electrons.

3 0

3 years ago

Question 7-Which of the following is not an acid-base neutralization reaction?

docker41 [41]

Answer:- It is choice D. $H_2SO_4+2NaCl\rightarrow 2HCl+Na_2SO_4$

Explanations:- In general, a neutralization reaction is the reaction of an acid with base to form salt and water.

in first reaction the reaction is taking place between ammonia(a base) and hydrochloric acid(an acid) to form their salt(ammonium chloride). So, it is an acid-base neutralization reaction.

In second reaction, sodium hydroxide(a base) is reacting with acetic acid(an acid) to form their salt(sodium acetate) and water. So, it is an acid-base neutralization reaction.

In third reaction, Nitric acid is reacting with calcium hydroxide(a base) to form a salt(calcium nitrate) and water. So, it is an acid-base neutralization reaction.

In fourth reaction, sulfuruc acid is reacting a sodium chloride(a salt) to give a double replacement reaction. It is not an acid-base neutralization reaction as it's not taking place between an acid and base.

So, the correct choice is D. $H_2SO_4+2NaCl\rightarrow 2HCl+Na_2SO_4$

7 0

3 years ago

Read 2 more answers

The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc

jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2HBr(g)\rightarrow H_2(g)+Br_2(g$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}$

Given: $-\frac{d[HBr]}{dt}]=0.301$

Putting in the values we get:

$\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}$

$+\frac{1d[Br_2]}{dt}=0.151$

Thus the rate of appearance of $Br_2$ is 0.151

8 0

3 years ago