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Contact [7]
2 years ago
6

Describe the structure of ammonium lauryl sulfate. Refer to the given diagram. Your answer should include the type of bonding, t

he elements contained, and the size and shape of the molecule. Write a short paragraph.

Chemistry
1 answer:
grandymaker [24]2 years ago
8 0

Answer:

Acidic inorganic salts, such as AMMONIUM LAURYL SULFATE, are generally soluble in water. The resulting solutions contain moderate concentrations of hydrogen ions and have pH's of less than 7.0. They react as acids to neutralize bases.

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The number of sp2 hybrid orbitals on the carbon atom in CO32– is
Vladimir79 [104]
The number of sp2 hybrid orbitals on the carbon atom in CO32– is 3. Because hybrids = combination of 2 different types of orbitals
sp2 = 1/3 s character + 2/3 p character
7 0
2 years ago
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How many valence electrons are in chlorodifluoromethane
Elis [28]
<span>Well... first, let's recognize that the chemical formula for chlorodifluoromethane is CHClF2. Count out how many valence electrons there are. C = 4, H = 1, Cl = 7, F (X2) = 14. Total is 26. Let's put C as the central atom, and put the other elements surrounding it. Draw a pair of electrons beach each element and the central atom. Then fill the halogen elements with 3 pairs of electrons each to fill their octets. Count out how many dots you have. There should be 26, making this the correct lewis structure! Remember, hydrogen doesn't have a full octet, only a maximum of two electrons.</span>
7 0
2 years ago
In science class, Blaine’s teacher puts one glow stick in a cup of hot water and another glow stick in a cup of cold water. She
timofeeve [1]

Answer:

The glow stick in hot water will be brighter

Explanation:

The glow stick in hot water will be brighter than the glow stick in cold water because the heat from the hot water will cause the molecules in the glow stick to move faster. The faster the molecules move in the glow stick, the sooner and brighter the reaction will be. The cold water will cause molecules to move slowly and it will take longer for the reaction to occur, which will also make it less bright.

3 0
2 years ago
Read 2 more answers
4. Convert the following: a. 4g mol of MgCl2 to g b. 2 lb mol of C3H8 to g c. 16 g of N2 to lb mol d. 3 lb of C2H6O to g mol
Nuetrik [128]

Answer:

a) 381.2 g

b) 39916 g

c) 0.0013 lb mol

d) 29.6 g mol

Explanation:

The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).

a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:

g mol = m/mw

4 = m/95.3

m = 381.2 g

b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.

lb mol = m/mw

2 = m/44

m = 88 lb

1 lb = 453.592 g

So, m = 88*453.592 = 39916 g

c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.

m = 16/453.592 = 0.0353 lb

lb mol = m/mw

lb mol = 0.0353/28

lb mol = 0.0013 lb mol

d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol

3 lb = 1360.78 g

g mol = m/mw

g mol = 1360.78/46

g mol = 29.6 g mol

6 0
2 years ago
Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl is
Rus_ich [418]

Answer:

0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

                M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

                no. of moles of NaCl = mass of NaCl / molar mass

                no. of moles of NaCl = 0.45g / 58.44 g/mol

               no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

                  M = moles of solute / Liter of solution . . . . . . . . . . (1)

                  M = 0.0077 g / 0.1 L

                  M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

3 0
2 years ago
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