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Mars2501 [29]
3 years ago
6

What is double bond,triple bond,lone pair and single bond

Chemistry
1 answer:
AlladinOne [14]3 years ago
5 0

In covalent bonds, atoms share electrons. These shared electrons are known as, well, covalent bonds!

In a covalent bond, each atom contributes one electron, meaning that for every bond, there are 2 electrons residing in it.

Definitions:

A single bond is a covalent bond that contains 2 electrons.

A double bond is a covalent bond that contains 4 electrons

A triple bond is a covalent bond that contains 6 electrons

It's very common for atoms to not use all of their electrons while bonding, which means that some will be left out and aren't used in any bonds. These are called lone pairs.

Definition:

A lone pair is a pair of electrons that aren't used in bonding, and therefore remain with its original atom.

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What is the bond energy required to break one mole of carbon-carbon bonds​
ArbitrLikvidat [17]

Answer:

<h2><u><em>100 kcal of bond energy</em></u></h2>

<u><em></em></u>

8 0
3 years ago
Choose one metal from the reactivity series that will react safely with dilute sulphuric acid
IceJOKER [234]

Answer:

Zinc (Zn) will safely react with dilute sulphuric acid.

Explanation:

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7 0
2 years ago
Explain the units and how to solve the following metric conversion then solve the metric conversion by filling in the blank 12KM
LiRa [457]

Answer:- 12 km = 12000 m

Solution:- It's a metric unit conversion where we are asked to convert 12 km to m where km stands for kilometer and m stands for meter.

In metric conversions, kilo means 1000.

So, 1 km = 1000 m

It means,  we multiply the given km by 1000 to get the answer in m as:

12km(\frac{1000m}{1km})

= 12000 m

Hence, 12 km = 12000 m.

5 0
3 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
3 years ago
What would A be? HELP ASAP
Minchanka [31]
Si has 4 available elections. Each Cl has 7.
7 x 4 = 28 + the 4 from your Si gives the total of 32 total electrons.
6 0
3 years ago
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