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lianna [129]
3 years ago
7

Which substance/s have covalent bond? Explain your answer.​

Chemistry
1 answer:
Viefleur [7K]3 years ago
4 0
Covalent bonding occurs when electrons are shared between atoms. Double and triple covalent bonds occur when four or six electrons are shared between two atoms, and they are indicated in Lewis structures by drawing two or three lines connecting one atom to another.
You might be interested in
(14.1.50) 0.03 mol of helium are enclosed in a cylinder with a piston which maintains a constant pressure in the helium. Initial
icang [17]

Answer:

a) 0.210 j /k

b) 0.032 j/k

Explanation:

Find the attachment for solution

3 0
3 years ago
Why is copper a pure substance? a. because it is an element b. because it is a compund c. because it is a homogenous mixture
Step2247 [10]

Answer:

a. because it is an element

Explanation:

its pure because it only has one type of atom, making it an element

6 0
2 years ago
Answers to all of these
Paul [167]

Answer:

1. Percent composition of  Al = 13.423 %

2.

  • Percent composition of Zn = 28.02 %
  • Percent composition of Cl = 30.6 %
  • Percent composition of O = 41.3 %

3. The empirical formula is C₅O₁₆

4. Molecular Formula= P₄O₆

Explanation:

Part first :

Data Given

Formula of the Molecule = Al₂ (CrO₄)₃

% of Al₂ = ?

> First of all find the atomic masses of each component in a molecule

For Al₂ (CrO₄)₃ atomic masses are given below

Al = 27 g/mol

Cr = 52 g/mol

O = 16 g/mol

> Then find the total masses of each component

2 atoms of Al = 27 g/mol x 2

= 54 g/mol

3 atoms of Cr = 52 g/mol x 3

= 156 g/mol

12 atoms of O = 16 g/mol x 12

= 192 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]

Molar Mass of Al₂ (CrO₄)₃ = 402

Now to find the mass percent of Al

Formula used to find the Mass percent of a component

Percent composition of  Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%

Put the values

Percent composition of  Al =  54 (g/mol) / 402 (g/mol) x 100%

Percent composition of  Al = 13.423 %

_______________________________________

Part 2

Data Given

Formula of the Molecule = Zn(ClO₃)₂

% Zn = ?

% Cl = ?

% O = ?

> First of all find the atomic masses of each component in a molecule

For Zn(ClO₃)₂ atomic masses are given below

Zn = 65 g/mol

Cl = 35.5 g/mol

O = 16 g/mol

> Then find the total masses of each component

1 atoms of Zn= 65 g/mol x 1

= 65 g/mol

2 atoms of Cl = 35.5 g/mol x  

= 71 g/mol

6 atoms of O = 16 g/mol x 6

= 96 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]

Molar Mass of Zn(ClO₃)₂ = 232g/mol

Now to find the mass percent of of each component one by one

1.  Formula used to find the mass percent of Zn

Percent composition of  Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%

Percent composition of Zn = 28.02 %

-------------------

2.  Formula used to find the mass percent of Cl

Percent composition of  Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%

Percent composition of Cl = 30.6 %

---------------------

3.  Formula used to find the mass percent of O

Percent composition of  O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%

Percent composition of O = 41.3 %

________________________________________

Part 3:

Data Given

Percentage of C = 27.3 %

Percentage of O = 72.7 %

Emperical Formula of the compound = ?

Solution:

So the compound has 27.3 % C and 72% O

First, find the mass of each of the elements in 100 g of the Compound.

C = 27.3 g

O = 72 g

Now find how many moles are there for each element in 100 g of compound

For this molar mass are required

That is

C = 12 g/mol

O = 16 g/mol

Formula Used

mole of C = mass of C / Molar mass of C

 mole of C = 27.3 / 12 g/mol

  mole of C = 2.275

Formula Used

mole of O = mass of O / Molar mass of O

 mole of O = 72g / 16 g/mol

  mole of O = 7.2

Divide each one by the smallest number of moles

C = 2.275 / 2.275

C = 1

O = 7.2 / 2.275

O = 3.2

Multiply the mole fraction to a number to get the whole number.

C = 1 x 5 = 5

O = 3.2 x 5 =  16

So, the empirical formula is C₅O₁₆

______________________________________

Part 4

Data Given

Percentage of P= 56.38 %

Percentage of O = 43.62%

Molar Mass = 219.9g

Molecular Formula of the compound = ?

Solution:

First, find the mass of each of the elements in 100 g of the Compound.

Mass of P= 56.38g

Mass of O = 43.62g

Now find how many moles are there for each element in 100 g of compound

find the moles in total compounds

Formula Used

mole of P = mass of  / Molar mass of P

 mole of P = 56.38 g / 31 g/mol

  mole of P = 1.818

Formula Used

mole of O = mass of O / Molar mass of O

 mole of O = 43. 62 / 16 g/mol

  mole of O = 2.7262

Now

first find the Emperical formula

Divide each one by the smallest number of moles

P = 1.818 /1.818

P= 1

for oxygen

O = 2.7262 / 1.818

O = 1.5

Multiply the mole fraction to a number to get the whole number.

P = 1 x 2 = 2

O = 1.5 x 2 =  3

So, the empirical formula is P₂O₃

Now  

Find molar mass of the empirical formula P₂O₃

2 (31) + 3 (16) = 62 + 48 = 110

Now find that how many empirical units are in a molecular unit.

(219.9 g/mol) / ( 110 g/mol) =  empirical units per molecular unit

empirical units per molecular unit = 1.999 =2

A here we get two empirical units in a molecular unit,

So the molecular formula is:

2 (P₂O₃) = P₄O₆

7 0
3 years ago
a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter
ladessa [460]

Answer:

The concentration of chloride ion is 2.82\times10^{-3}\;mol/L

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the Cl^- content 100 ppm suggests the presence of 100 mg of Cl^- in 1 L of solution.

The molar mass of Cl^- is equal to the molar mass of Cl atom as the mass of the excess electron in Cl^- is negligible as compared to the mass of Cl atom.

So, the molar mass of Cl^- is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of Cl^- present in 100 mg (0.100 g) of Cl^- is calculated as shown below:

N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol

So, there is 2.82\times10^{-3}\;mol of Cl^- present in 1 L of solution.

5 0
3 years ago
Draw the condensed structural formula for the ester formed when each of the following reacts with methyl alcohol:
Vikki [24]

Answer:

1) HCOOCH3

2) CH3CH2COOCH3

3) CH3CH2CH2CH(CH3)COOCH3

Explanation:

In the reaction between an alcohol and a carboxylic acid, an ester and water are formed. It is analogous to the inorganic neutralization reaction but this reaction is called esterification in organic chemistry. Esters contain the general formula RCOOR where the RCOO moiety was obtained from the acid and the other R moiety was obtained from the alcohol. The -COOR shows the ester linkage in the molecule. The condensed structural formulas shown in the answer reflects these facts.

6 0
3 years ago
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