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tino4ka555 [31]
3 years ago
13

How many protons are there in 20.02 mol of neon (ne)?

Chemistry
2 answers:
hodyreva [135]3 years ago
7 0
In order to determine the number of protons in 20.02 mol of Ne, we use Avogadro's number to convert the number of moles to number of atoms, 1 mol = 6.022 x 10^23 atoms. From there, we must know the number of protons in a Neon atom, which is 10. Thus, the formula will be:

(20.02 mol Ne)x(6.022 x 10^23 atoms/mol)x(10 protons/1 atom Ne) =
1.2056 x 10^26 protons 
Tanya [424]3 years ago
5 0

Answer: There are 1.216×10^26 protons in 20.02mole of neon

Explanation:

Using Avogadro's number

1 mole of neon contains 6.023×10^23 neon atoms.

I atm contains 10 protons

Number of protons = (6.023×10^23)×20.02×10= 1.216×10^26

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Answer:

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Explanation:

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What will increased temperature in a reaction cause
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I think your answer is 2 because the temperature will rise so the particles will move faster
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Draw the bridged bromonium ion that is formed as an intermediate during the bromination of this alkene. include hydrogen atoms,
gogolik [260]
<h2>Answer</h2>

Bromination:

Any reaction or process in which bromine (and no other elements) are introduced into a molecule.

Bromonium Ion:

The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)

Mechanism:

Step 1:

In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".

Step 2:

When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.

In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.

6 0
3 years ago
0.01040 m as an integer
SpyIntel [72]

Answer:

Here,

0.01040 m as an integer= 1.04 × 10-2

5 0
2 years ago
Sort each of the following elements by effective nuclear charge, Zeff, from smallest to largest:
krek1111 [17]

Answer:

Rb = +1 , Sr = +2, In= +3,  Sn = +4, Sb= +5

Explanation:

Formula:

Zeff = Z - S

Z = atomic number

S = number of core shell or inner shell electrons

For Sn:

Electronic configuration:

Sn₅₀ = [Kr] 4d¹⁰ 5s² 5p²

Zeff = Z - S

Zeff = 50 - 46

Zeff = +4

For Rb:

Electronic configuration:

Rb₃₇ = [Kr] 5s¹

Zeff = Z - S

Zeff = 37 - 36

Zeff = +1

For Sb:

Electronic configuration:

Sb₅₁ = [Kr] 4d¹⁰ 5s² 5p³

Zeff = Z - S

Zeff = 51 - 46

Zeff = +5

For In:

Electronic configuration:

In₄₉ = [Kr] 4d¹⁰ 5s² 5p¹

Zeff = Z - S

Zeff = 49 - 46

Zeff = +3

For Sr:

Electronic configuration:

Sr₃₈= [Kr]  5s²

Zeff = Z - S

Zeff = 38 - 36

Zeff = +2

8 0
3 years ago
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