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stiv31 [10]
4 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]4 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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Calculate the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, given that the Ksp of Ca(OH)2 is 4.96 x 10-6 at that tem
Ghella [55]

The solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, at given Ksp of Ca(OH)2 is determined as 1.626 mM.

<h3>Dissociation reaction of Ca(OH)2</h3>

The dissociation reaction of Ca(OH)2 is given as follows;

Ca(OH)₂  ⇄ Ca²⁺  + 2OH⁻¹

                     x           2x

Concentration of Ca²⁺ = 0.469 M

Ksp = [x][2x]²

ksp = (0.469)(2x²)

ksp = 4(0.469)x²

ksp = 1.876x²

4.96 x 10⁻⁶ = 1.876x²

x² = (4.96 x 10⁻⁶)/(1.876)

x² = 2.643 x 10⁻⁶

x = √(2.643 x 10⁻⁶)

x = 1.626 x 10⁻³ M

x = 1.626 mM

Thus, the solubility of Ca(OH)2 in a 0.469M CaCl2 solution at 31°C, at given Ksp of Ca(OH)2 is determined as 1.626 mM.

Learn more about solubility here: brainly.com/question/23946616

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3 0
2 years ago
24. As the first five elements in Group 14 are considered in order from top to bottom, there are changes
myrzilka [38]

Answer:

B

Explanation:

As you move down the periodic table in the carbon family, the atomic radius and ionic radius increase while electronegativity and ionization energy decrease. Atom size increases moving down the group because an additional electron shell is added.

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Which fats result when manufacturers add hydrogen to fat molecules in vegetable oil?
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3 years ago
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A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper
finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1:  Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL *  0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene  = 35.146 grams / 78 g/mol  = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4      = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

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