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stiv31 [10]
3 years ago
5

Are the bonds that chlorine forms with sodium (to form NaCl) and with carbon (to form CCl4) the same in both compounds?

Chemistry
1 answer:
s2008m [1.1K]3 years ago
3 0

Answer:

Organic chemistry is very important to know the strength and forms of the compounds.

Explanation:

There are two extreme bonding that is ionic and covalent. There are some ionic bonds that contains the covalent bonding. Some of ionic bond are partially covalent and some covalent bond are ionic. Polar covalent have the extreme type of bonding.

Most of the carbon compound are bonded covalently. But these are partial ionic too. Polarity is defined as the measurement of the separation of the charges of the compounds at both the ends.

Ionic compounds allow synthesis in organic compound. Covalent bonds are differently important for carbon molecules bonding.

Thus the bonds that chlorine forms with sodium is different and when form with carbon then compounds will be different

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How any moles of H2O will be produced from 12.3 moles of HCL reacting with Ca(OH)2?
stiks02 [169]

Answer:

\large \boxed{\text{12.3 mol HCl}}  

Explanation:

We need a balanced chemical equation with moles.

            2HCl +Ca(OH)₂ ⟶ CaCl₂ + 2H₂O

n/mol:    12.3

The molar ratio is 2 mol H₂O:2 mol HCl.

\text{Moles of H$_{2}$O} = \text{12.3 mol HCl} \times \dfrac{\text{2 mol H$_{2}$O}}{\text{2 mol HCl}} = \textbf{12.3 mol HCl}\\\\\text{The reaction produces $\large \boxed{\textbf{12.3 mol HCl}}$}

3 0
3 years ago
Bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (so2) exhibits
Zepler [3.9K]

Bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (SO₂) exhibits a dipole moment while carbon dioxide (CO₂) does not because of the difference in their shape, CO₂ is having linear geometry thus exhibit zero dipole moment while SO₂ is having bent shape thus exhibit dipole moment. So, despite the fact that bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (SO₂) exhibits a dipole moment while carbon dioxide (CO₂) does not.

3 0
3 years ago
Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati
worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4  +  Cl-

Firstly, we calculate the number of moles of  the ammonia  as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

                 NH4+      H2O     ⇄  NH3        H3O+

I                0.242                           0             0

C                 -X                              +x              +X

E             0.242-X                          X              X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b    = K w

 

, where  K w- the self-ionization constant of water, equal to  

10 ^-14  at room temperature

This means that you have

K a = K w.K b   = 10 ^− 14 /1.8 * 10^-5 =  5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈  0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0
3 years ago
What method is used to removed the unreacted copper (II) carbonate?​
Evgen [1.6K]

Answer:

Filtration

Explanation:

Metal carbonate is insoluble, it is possible to filter off the unreacted substances leaving the desired salt solution

8 0
3 years ago
How many moles are in 1.51 x 10^24 molecules of water?
lyudmila [28]

Answer:One mole of HBr has 6.02 x 1

0

23

molecules of HBr.

1 mole of HBr = 6.02 x 1

0

23

molecules of HBr.-----(a)

X mole of HBr has 1.21 x

10

24

molecules of HBr.

X mole of HBr = 1.21 x

10

24

molecules of HBr------(b)

Taking ratio of (a) and (b)

X / 1 = 1.21 x

10

24

/ 6.02 x 1

0

23

X= 2.009 moles.

Explanation:

3 0
3 years ago
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